A Mystical Whole

1 of 102 A Mystical Whole By Ian Beardsley © 2016

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Object Status Ian Beardsley 2016 

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4 of 102 #include <stdio.h> int main (void) { printf("\n"); float object, Tzero, T, time; int n; printf("If we use alphacentauri as the key to our model,\n"); printf("for modeling the future, then our task has been reduced,\n"); printf("through the work I have done, to quite a simple one.\n"); printf("growthrate=k=0.0621, objective=log 100/log e = 4.6 achievements,\n"); printf("Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.0.12(4.6)\n"); printf("1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age of gold:silver\n"); printf("Our equation is then, Time=(Object Achieved)/(Achievements/ year)\n"); printf("\n"); do { printf("How many simulations would you like to run (10 max)? "); scanf("%d", &n); } while (n>10 && n<=0); for (int i=1; i<=n; i++) { do { printf("What is distance to object(0-4.6?) "); scanf("%f", &object); } while (n<=0 && n>=4.6); printf("What is the starting point (year) ?"); scanf("%f", &Tzero); T= object/(0.0621); time= Tzero+T; printf("Time to object=%f years.\n", T); printf("That is the year: %f\n", time); printf("\n"); } }

5 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./object If we use alphacentauri as the key to our model, for modeling the future, then our task has been reduced, through the work I have done, to quite a simple one. growthrate=k=0.0621, objective=log 100/log e = 4.6 achievements, Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.0.12(4.6) 1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age of gold:silver Our equation is then, Time=(Object Achieved)/(Achievements/year) How many simulations would you like to run (10 max)? 10 What is distance to object(0-4.6?) 0.25 What is the starting point (year) ?1969 Time to object=4.025765 years.

That is the year: 1973.025757 What is distance to object(0-4.6?) 0.50 What is the starting point (year) ?1969 Time to object=8.051530 years.

That is the year: 1977.051514 What is distance to object(0-4.6?) 0.75 What is the starting point (year) ?1969 Time to object=12.077294 years.

That is the year: 1981.077271 What is distance to object(0-4.6?) 1.00 What is the starting point (year) ?1969 Time to object=16.103060 years.

That is the year: 1985.103027 What is distance to object(0-4.6?) 1.25 What is the starting point (year) ?1969 Time to object=20.128824 years.

That is the year: 1989.128784 What is distance to object(0-4.6?) 1.50 What is the starting point (year) ?1969 Time to object=24.154589 years.

That is the year: 1993.154541 What is distance to object(0-4.6?) 1.75 What is the starting point (year) ?1969 Time to object=28.180355 years.

That is the year: 1997.180298 What is distance to object(0-4.6?) 2.00 What is the starting point (year) ?1969

6 of 102 Time to object=32.206120 years.

That is the year: 2001.206177 What is distance to object(0-4.6?) 2.25 What is the starting point (year) ?1969 Time to object=36.231884 years.

That is the year: 2005.231934 What is distance to object(0-4.6?) 2.50 What is the starting point (year) ?1969 Time to object=40.257648 years.

That is the year: 2009.257690 jharvard@appliance (~/Dropbox): â&#x20AC;Š

7 of 102 Last login: Tue Apr 5 20:29:12 on ttys000 /Users/ianbeardsley/Desktop/object\ run ; exit; Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/object\ run ; exit; If we use alphacentauri as the key to our model, for modeling the future, then our task has been reduced, through the work I have done, to quite a simple one. growthrate=k=0.0621, objective=log 100/log e = 4.6 achievements, Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.0.12(4.6) 1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age of gold:silver Our equation is then, Time=(Object Achieved)/(Achievements/year) How many simulations would you like to run (10 max)? 4 What is distance to object(0-4.6?) 4 What is the starting point (year) ?1969 Time to object=64.412239 years.

That is the year: 2033.412231 What is distance to object(0-4.6?) 0 What is the starting point (year) ?1969 Time to object=0.000000 years.

That is the year: 1969.000000 What is distance to object(0-4.6?) 4.6 What is the starting point (year) ?1969 Time to object=74.074074 years.

That is the year: 2043.074097 What is distance to object(0-4.6?) 4.5 What is the starting point (year) ?1969 Time to object=72.463768 years.

That is the year: 2041.463745 logout [Process completed]

8 of 102 Last login: Tue Apr 5 20:29:12 on ttys000 /Users/ianbeardsley/Desktop/object\ run ; exit; Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/object\ run ; exit; If we use alphacentauri as the key to our model, for modeling the future, then our task has been reduced, through the work I have done, to quite a simple one. growthrate=k=0.0621, objective=log 100/log e = 4.6 achievements, Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.

0.12(4.6) 1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age of gold:silver Our equation is then, Time=(Object Achieved)/(Achievements/year) How many simulations would you like to run (10 max)? 4 What is distance to object(0-4.6?) 4 What is the starting point (year) ?1969 Time to object=64.412239 years.

That is the year: 2033.412231 What What Time That

is is to is

distance to object(0-4.6?) 0 the starting point (year) ?1969 object=0.000000 years. the year: 1969.000000

What What Time That

is is to is

distance to object(0-4.6?) 4.6 the starting point (year) ?1969 object=74.074074 years. the year: 2043.074097

What What Time That

is is to is

distance to object(0-4.6?) 4.5 the starting point (year) ?1969 object=72.463768 years. the year: 2041.463745

logout [Process completed]â&#x20AC;Š

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11 of 102 In an effort to build a base for those unknowables that have been seemingly imposable to understand, such as will an asteroid come from deep space on a collision course with earth with little warning to act, or, in general, will humans do the wrong thing for their survival because of what they cannot seemingly know, the author has been set since the year 2000 on an intensive campaign to learn what he can is such an area owing to the idea that his life path has been so far off the trodden path, that he can access some of these things to a certain degree, if he remains unseen in his endeavors. The project has been met with an extraordinary success, and these are the things he has found to date April 2016, as a starting point.

Ian Beardsley April 03, 2016â&#x20AC;Š

12 of 102 It will be necessary to tell two stories for the handwritten numbers and equations that follow to make sense to anyone.â&#x20AC;Š

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Ian Beardsley April 02 2016 9:00 PMâ&#x20AC;Š

14 of 102 Story One: Manuel Herediaâ&#x20AC;Š

15 of 102 Gypsy Shamanism And The Universe I wrote a short story last night, called Gypsy Shamanism and the Universe about the AE-35 unit, which is the unit in the movie and book 2001: A Space Odyssey that HAL reports will fail and discontinue communication to Earth. I decided to read the passage dealing with the event in 2001 and HAL, the ship computer, reports it will fail in within 72 hours. Strange, because Venus is the source of 7.2 in my Neptune equation and represents failure, where Mars represents success.

Ian Beardsley August 5, 2012â&#x20AC;Š

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Chapter One It must have been 1989 or 1990 when I took a leave of absence from The University Of Oregon, studying Spanish, Physics, and working at the state observatory in Oregon -- Pine Mountain Observatory—to pursue flamenco in Spain.

The Moors, who carved caves into the hills for residence when they were building the Alhambra Castle on the hill facing them, abandoned them before the Gypsies, or Roma, had arrived there in Granada Spain. The Gypsies were resourceful enough to stucco and tile the abandoned caves, and take them up for homes.

Living in one such cave owned by a gypsy shaman, was really not a down and out situation, as these homes had plumbing and gas cooking units that ran off bottles of propane. It was really comparable to living in a Native American adobe home in New Mexico.

Of course living in such a place came with responsibilities, and that included watering its gardens. The Shaman told me: “Water the flowers, and, when you are done, roll up the hose and put it in the cave, or it will get stolen”. I had studied Castilian Spanish in college and as such a hose is “una manguera”, but the Shaman called it “una goma” and goma translates as rubber. Roll up the hose and put it away when you are done with it: good advice!

So, I water the flowers, rollup the hose and put it away. The Shaman comes to the cave the next day and tells me I didn’t roll up the hose and put it away, so it got stolen, and that I had to buy him a new one.

He comes by the cave a few days later, wakes me up asks me to accompany him out of The Sacromonte, to some place between there and the old Arabic city, Albaicin, to buy him a new hose.

It wasn’t a far walk at all, the equivalent of a few city blocks from the caves. We get to the store, which was a counter facing the street, not one that you could enter. He says to the man behind the counter, give me 5 meters of hose. The man behind the counter pulled off five meters of hose from the spindle, and cut the hose to that length. He stated a value in pesetas, maybe 800, or so, (about eight dollars at the time) and the Shaman told me to give that amount to the man behind the counter, who was Spanish. I paid the man, and we left.

I carried the hose, and the Shaman walked along side me until we arrived at his cave where I was staying. We entered the cave stopped at the walk way between living room and kitchen, and he said: “follow me”. We went through a tunnel that had about three chambers in the cave, and entered one on our right as we were heading in, and we stopped and before me was a collection of what I estimated to be fifteen rubber hoses sitting on ground. The Shaman told me

17 of 102 to set the one I had just bought him on the floor with the others. I did, and we left the chamber, and he left the cave, and I retreated to a couch in the cave living room.â&#x20AC;Š

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Chapter Two Gypsies have a way of knowing things about a person, whether or not one discloses it to them in words, and The Shaman was aware that I not only worked in Astronomy, but that my work in astronomy involved knowing and doing electronics.

So, maybe a week or two after I had bought him a hose, he came to his cave where I was staying, and asked me if I would be able to install an antenna for television at an apartment where his nephew lived.

So this time I was not carrying a hose through The Sacromonte, but an antenna.

There were several of us on the patio, on a hill adjacent to the apartment of The Shaman’s Nephew, installing an antenna for television reception.

Chapter Three I am now in Southern California, at the house of my mother, it is late at night, she is a asleep, and I am about 24 years old and I decide to look out the window, east, across The Atlantic, to Spain. Immediately I see the Shaman, in his living room, where I had eaten a bowl of the Gypsy soup called Puchero, and I hear the word Antenna. I now realize when I installed the antenna, I had become one, and was receiving messages from the Shaman.

The Shaman’s Children were flamenco guitarists, and I learned from them, to play the guitar. I am now playing flamenco, with instructions from the shaman to put the gypsy space program into my music. I realize I am not just any antenna, but the AE35 that malfunctioned aboard The Discovery just before it arrived at the planet Jupiter in Arthur C. Clarke’s and Stanley Kubrick’s “2001: A Space Odyssey”. The Shaman tells me, telepathically, that this time the mission won’t fail.

Chapter Four I am watching Star Wars and see a spaceship, which is two oblong capsules flying connected in tandem. The Gypsy Shaman says to me telepathically: “Dios es una idea: son dos”. I understand that to mean “God is an idea: there are two elements”. So I go through life basing my life on the number two.

Chapter Five Once one has tasted Spain, that person longs to return. I land in Madrid, Northern Spain, The Capitol. The Spaniards know my destination is Granada, Southern Spain, The Gypsy

19 of 102 Neighborhood called The Sacromonte, the caves, and immediately recognize I am under the spell of a Gypsy Shaman, and what is more that I am The AE35 Antenna for The Gypsy Space Program. Flamenco being flamenco, the Spaniards do not undo the spell, but reprogram the instructions for me, the AE35 Antenna, so that when I arrive back in the United States, my flamenco will now state their idea of a space program. It was of course, flamenco being flamenco, an attempt to out-do the Gypsy space program.

Chapter Six I am back in the United States and I am at the house of my mother, it is night time again, she is asleep, and I look out the window east, across the Atlantic, to Spain, and this time I do not see the living room of the gypsy shaman, but the streets of Madrid at night, and all the people, and the word Jupiter comes to mind and I am about to say of course, Jupiter, and The Spanish interrupt and say â&#x20AC;&#x153;Yes, you are right it is the largest planet in the solar system, you are right to consider it, all else will flow from it.â&#x20AC;?

I know ratios, in mathematics are the most interesting subject, like pi, the ratio of the circumference of a circle to its diameter, and the golden ratio, so I consider the ratio of the orbit of Saturn (the second largest planet in the solar system) to the orbit of Jupiter at their closest approaches to The Sun, and find it is nine-fifths (nine compared to five) which divided out is one point eight (1.8).

I then proceed to the next logical step: not ratios, but proportions. A ratio is this compared to that, but a proportion is this is to that as this is to that. So the question is: Saturn is to Jupiter as what is to what? Of course the answer is as Gold is to Silver. Gold is divine; silver is next down on the list. Of course one does not compare a dozen oranges to a half dozen apples, but a dozen of one to a dozen of the other, if one wants to extract any kind of meaning. But atoms of gold and silver are not measured in dozens, but in moles. So I compared a mole of gold to a mole of silver, and I said no way, it is nine-fifths, and Saturn is indeed to Jupiter as Gold is to Silver.

I said to myself: How far does this go? The Shamanâ&#x20AC;&#x2122;s son once told me he was in love with the moon. So I compared the radius of the sun, the distance from its center to its surface to the lunar orbital radius, the distance from the center of the earth to the center of the moon. It was Nine compared to Five again!

Chapter Seven I had found 9/5 was at the crux of the Universe, but for every yin there had to be a yang. Nine fifths was one and eight-tenths of the way around a circle. The one took you back to the beginning which left you with 8 tenths. Now go to eight tenths in the other direction, it is 72 degrees of the 360 degrees in a circle. That is the separation between petals on a five-petaled flower, a most popular arrangement. Indeed life is known to have five-fold symmetry, the physical, like snowflakes, six-fold. Do the algorithm of five-fold symmetry in reverse for six-fold symmetry, and you get the yang to the yin of nine-fifths is five-thirds.

20 of 102 Nine-fifths was in the elements gold to silver, Saturn to Jupiter, Sun to moon. Where was fivethirds? Salt of course. “The Salt Of The Earth” is that which is good, just read Shakespeare’s “King Lear”. Sodium is the metal component to table salt, Potassium is, aside from being an important fertilizer, the substitute for Sodium, as a metal component to make salt substitute.

The molar mass of potassium to sodium is five to three, the yang to the yin of nine-fifths, which is gold to silver. But multiply yin with yang, that is nine-fifths with five-thirds, and you get 3, and the earth is the third planet from the sun.

I thought the crux of the universe must be the difference between nine-fifths and five-thirds. I subtracted the two and got two-fifteenths! Two compared to fifteen! I had bought the Shaman his fifteenth rubber hose, and after he made me into the AE35 Antenna one of his first transmissions to me was: “God Is An Idea: There Are Two Elements”.

It is so obvious, the most abundant gas in the Earth Atmosphere is Nitrogen, chemical group 15 and the Earth rotates through 15 degrees in one hour.

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Story Two: Paul Levinsonâ&#x20AC;Š

22 of 102 I have written three papers on the anomaly of how my scientific investigation shows the Universe related to the science fiction of Paul Levinson, Isaac Asimov, and Arthur C. Clarke. In my last paper, “The Levinson-Asimov-Clarke Equation” part of the comprehensive work “The Levinson, Asimov, Clarke Triptic, I suggest these three authors should be taken together to make some kind of a whole, that they are intertwined and at the heart of science fiction. I have now realized a fourth paper is warranted, and it is just the breakthrough I have been looking for to put myself on solid ground with the claim that fiction is related to reality in a mathematical way pertaining to the Laws of Nature. I call it Fiction-Reality Entanglement.

In my paper Paul Levinson, Isaac Asimov, Arthur C. Clarke Intertwined With An Astronomer’s Research, I make the mathematical prediction that “humans have a 70% chance of developing Hyperdrive in the year 2043” to word it as Paul Levinson worded it, and I point out that this is only a year after the character Sierra Waters is handed a newly discovered document that sets in motion the novel by Paul Levinson, “The Plot To Save Socrates”.

I now find that Isaac Asimov puts such a development in his science fiction at a similar time in the future, precisely in 2044, only a year after my prediction and two years after Sierra Waters is handed the newly discovered document that initiates her adventure. So, we have my prediction, which is related to the structure of the universe in a mystical way right in between the dates of Levinson and Asimov, their dates only being a year less and a year greater than mine.

Asimov places hyperdrive in the year 2044 in his short story “Evidence” which is part of his science fiction collection of short stories called, “I, Robot”.

This is a collection of short stories where Robot Psychologist Dr. Susan Calvin is interviewed by a writer about her experience with the company on earth in the future that first developed sophisticated robots. In this book, the laws of robotics are created and the idea of the positronic brain introduced, and the nature of the impact robots would have on human civilization is explored. Following this collection of stories Asimov wrote three more novels, which comprise the robot series, “The Caves of Steel”, “The Naked Sun”, and “The Robots of Dawn”.

“I, Robot” is Earth in the future just before Humanity settles the more nearby stars. The novels comprising “The Robot Series” are when humanity has colonized the nearby star systems, The Foundation Trilogy, and its prequels and sequels are about the time humanity has spread throughout the entire galaxy and made an Empire of it. All of these books can be taken together as one story, with characters and events in some, occurring in others.

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Hyperdrive is invented in I, Robot by a robot called The Brain, owned by the company for which Dr. Susan Calvin works when it is fed the mathematical logistical problems of making hyperdrive, and asked to solve them. It does solve them and it offers the specs on building an interstellar ship, for which two engineers follow in its construction. They are humorously sent across the galaxy by The Brain, not expecting it, and brought back to earth in the ship after they constructed it. This was in the story in “I, Robot” titled “Escape!”.

But Dr. Susan Calvin states in the following short story, that I mentioned, “Evidence”: “But that wasn’t it, either”…”Oh, eventually, the ship and others like it became government property; the Jump through hyperspace was perfected, and now we actually have human colonies on the planets of some of the nearer stars, but that wasn’t it.”

“It was what happened to the people here on Earth in last fifty years that really counts.”

And, what happened to people on Earth? The answer is in the same story “Evidence” from which that quote is at the beginning. It was when the Regions of the Earth formed The Federation. Dr. Susan Calvin says at the end of the story “Evidence”: “He was a very good mayor; five years later he did become Regional Co-ordinator. And when the Regions of Earth formed their Federation in 2044, he became the first World Co-ordinator.”

It is from that statement that I get my date of 2044 as the year Asimov projects for hyperdrive.

Ian Beardsley March 17, 2011

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I watched a video on youtube about Terence McKenna where he lectured on his timewave zero theory. I found there was not an equation for his timewave zero graph but that a computer algorithm generated the graph of the wave. The next day I did a search on the internet to see if a person could download timewave software for free. As it turned out one could, for both Mac and pc. It is called â&#x20AC;&#x153;Timewave Calculator Version 1.0â&#x20AC;?. I downloaded the software and found you had to download it every time after you quit the application and that you could not save the graph of your results or print them out. So I did a one-time calculation. It works like this: you input the range of time over which you want see the timewave and you cannot calculate past 2012, because that is when the timewave ends. You also put in a target date, the time when you want to get a rating for the novelty of the event that occurred on that day. You can also click on any point in the graph to get the novelty rating for that time. I put in:

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Input: Begin Date: December 27 1968 18 hours 5 minutes 37 seconds End Date: December 2 2011 0 hours 28 minutes 7 seconds McKenna said in the video on youtube that the dips, or valleys, in the timewave graph represent novelties. So, I clicked on the first valley after 1969 since that is the year we went to the moon, and the program gave its novelty as: Sheliak Timewave Value For Target: 0.0621 On Target Date: August 4, 1969 9 hours 53 minutes 38 seconds I was happy to see this because, I determined that the growth rate constant, k, that rate at which we progress towards hyperdrive, in my calculation in my work Asimovian Prediction For Hyperdrive, that gave the date 2043, a year after Sierra Waters was handed the newly discovered document that started her adventure in The Plot To Save Socrates, by Paul Levinson, and a year before Isaac Asimov had placed the invention of hyperdrive in his book I, Robot, was: (k=0.0621) The very same number!!!

What does that mean? I have no idea; I will find out after I buy The Invisible Landscape by Terence McKenna, Second Edition, and buy a more sophisticated timewave software than that which is offered for free on the net.

Ian Beardsley March 19, 2011

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There is a common thread running through the Science Fiction works of Paul Levinson, Isaac Asimov, and Arthur C. Clarke.

In the case of Isaac Asimov, we are far in the future of humanity. In his Robot Series, Asimov has man making robots whose programming only allows them to do that which is good for humanity.

As a result, these robots, artificial intelligence (AI), take actions that propel humanity into settling the Galaxy, in the robot series, and ultimately save humanity after they have settled the Galaxy and made an empire of it (In the Foundation Series).

In the case of Paul Levinson, scholars in the future travel through time and use cloning, a concept related to artificial intelligence (it is the creating of human replicas as well, but biological, not electronic), and the goal is to save great ancient thinkers from Greece, and to manipulate events in the past for a positive outcome for the future of humanity, just as the robots try to do in the work of Asimov.

In the case of Arthur C. Clarke, man undergoes a transformation due to a monolith placed on the moon and earth by extraterrestrials who have created life on earth. The monolith is a computer. It takes humans on a voyage to other planets in the solar system, and in their trials, humanity goes through trials that result in a transformation for the ending of their dependence on their technology and for becoming adapted to life in the Universe beyond Earth. That is, the character Dave Bowman becomes the Starchild in his mission to Jupiter. The artificial intelligence is the ship computer called HAL. So, the thread is the salvation of man through technology, and their transformation to a new human paradigm, where they can end their dependence on Earth and adapt to the nature of the Universe as a whole.

At the time I was reading these novels, I was doing astronomical research, and, to my utter astonishment, my relationships I was discovering pertaining to the Universe were turning up times and values pivotal to these works of Levinson, Asimov, and Clarke. Further, I was interpreting much of my discoveries by developing them in the context of short fictional stories.

In my story, â&#x20AC;&#x153;The Questionâ&#x20AC;?, we find Artificial Intelligence is in sync with the phases of the first appearance of the brightest star Sirius for the year, and the flooding of the Nile river, which brings in the Egyptian agricultural season. It is presumed by some scholars that because the Egyptian calendar is in sync with the Nile-Sirius cycle, theirs began four such cycles ago.

I then relate that synchronization to another calculation that turns up the time when the key figure of the Foundation Series of Asimov begins his program to found a civilization that will save the galaxy. We later find his actions were manipulated into being by robots, in order to save intelligent life in the galaxy by creating a viable society for it called Galaxia.

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In the case of Paul Levinson, I was making a calculation to predict when man would develop hyperdrive, that engine which could take us to the stars, and end our dependence on an Earth that cannot take care of humans forever. That time turned out to be when the key scholar in the work

by Paul Levinson, began her quest to help humanity by traveling into the past and using cloning, in part, to change history for the better. I can now only feel her quest to save humanity is going to be through changing history to bring about the development of hyperdrive, so humanity will no longer depend on Earth alone, which, as I have said, cannot take care of life forever.

Finally, where Arthur C. Clarke is concerned, I find values in the solar system and nature that are in his monolith, and I connect it to artificial intelligence of a sort, that kind which would be based on silicon.

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I had tried to predict mathematically when we would develop hyperdrive, and it came out just a year after the character, Sierra Waters, in the science fiction piece by Paul Levinson titled â&#x20AC;&#x153;The Plot To Save Socratesâ&#x20AC;? was handed a newly discovered document at the beginning of the book that got the whole story rolling. I wrote in my piece â&#x20AC;&#x153;Forecast For Hyperdrive: A Study In Asimovian Psychohistory: It is a curious thing that the Earth is the third planet from the Sun and the third brightest star in the sky is the closest to us and very similar to the sun in a galaxy of a rich variety of stars. This closest star to us is a triple system known as Alpha Centauri A, B, and Proxima Centauri. Alpha Centauri A is, like our Sun a main sequence spectral type G star. Precisely, G2 V, just as is the Sun. Its physical characteristics are very close to those of the Sun: 1.10 solar masses, 1.07 solar diameters, and 1.5 solar luminosities. It is absolute magnitude +4.3. The absolute magnitude of the Sun is +4.83.

If ever the option existed for humans to travel to the stars, this situation speaks of it, whether or not Alpha Centauri has an earth-like planet in its habitable zone.

It has been said that the base ten place significant system of writing numbers stems from the fact we have ten fingers to count on. In so far as science can save us, it can destroy us in that science is not dangerous, but humans can be.

Traveling to the planets is possible with chemical fuel rockets, but traveling to the stars is another story, because of their immense distances from us, and from one another.

What are the odds that our development in technologies will take us to the stars before we destroy ourselves first? In other words, what are the intrinsic odds for humankind to develop the hyperdrive before without bringing about its own end first?

We do a random walk to Alpha Centauri of 10 one light year jumps. We make 10 equal steps randomly of one light year each, equal steps that if all are towards Alpha Centauri we will land beyond it. If 10 are away from it, we are as far from it as can be. And, if 5 are towards it, and five are away from it, we have gone nowhere.

In this allegory we calculate the probability of landing on Alpha Centauri, in 10 random leaps of a light year each, a light year being the distance light travels in the time it takes the earth to make one revolution around the sun, light speed a natural constant. (continue to the next page)

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!

The probability of making n steps in either direction forms a bell shaped curve. After 10 randomly made steps the odds of going nowhere is highest and, is represented by five in the bell curve corresponding to 0. Let us round the distance of Alpha Centauri to four light years, giving humans the benefit of the doubt. The number positive four in the bell graph has written above it the number 7. Seven out of ten times 100 for effort gives a 70% chance of making it to the stars without becoming extinct first. I believe the percent understanding of our technological development towards hyperdrive, where we have just entered space with chemical rockets and developed fast, compact, computers, is given by:

!

Evaluated at n1=7.

N is 10 steps.

And n1 is the number of steps towards Alpha Centauri, n2 those away from it.

And, p is the probability that the step is towards Alpha Centauri, and q is the probability that the step is away from Alpha Centauri.

N = n1 + n2 And m = n1-n2 is the displacement And q+p=1â&#x20AC;Š

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The trick to using this equation is in knowing the possible combinations of steps that can be made that equal 10. Like five right, five left with a displacement of 0 or, 10 right, 0 left with a displacement of 10 or, 7 left, 3 right with a displacement of negative 4.

To land at 4 light years from earth, with 10 one light year jumps, one must go away from Alpha Centauri 3 jumps of a light year each then 7 jumps toward it of one light year each, to land on it, that is to land at +4, its location. So n1 is 7 and n2 is 3. The probability to jump away from the star is 1/2 and the probability to jump towards it is 1/2. That is p=1/2 and q=1/2. There are ten random jumps, so, N=10.

Using our equation:

!

We would be, by this reasoning 12% along in the development towards hyperdrive.

Ian Beardsley June 2009 If human technology has ever been anything, it has been exponential, growing in proportion to itself. In other words, two developments beget 8, eight beget 16, and sixteen begets 32. My grandfather rode a horse when he was a child, as a young man he drove a car, and when I knew him as a child, he saw humans land on the moon.

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It wasnâ&#x20AC;&#x2122;t long before we made computers small enough that people could keep in their homes that did more than computers did in the 60â&#x20AC;&#x2122;s that filled an entire room.

Having calculated that we are 12% along in developing the hyperdrive, we can use the equation for natural growth to estimate when we will have hyperdrive. It is of the form: ! t is time and k is a growth rate constant which we must determine to solve the equation. In 1969 Neil Armstrong became the first man to walk on the moon. In 2009 the European Space Agency launched the Herschel and Planck telescopes that will see back to near the beginning of the universe. 2009-1969 is 40 years. This allows us to write: ! log 12 = 40k log 2.718 0.026979531 = 0.4342 k k=0.0621 We now can write: ! ! log 100 = (0.0621) t log e t = 74 years 1969 + 74 years = 2043 Our reasoning would indicate that we will have hyperdrive in the year 2043.

Study summary: 1. We have a 70% chance of developing hyperdrive without destroying ourselves first.

2. We are 12% along the way in development of hyperdrive.

3. We will have hyperdrive in the year 2043, plus or minus.

32 of 102 Sierra Waters was handed the newly discovered document in 2042.â&#x20AC;Š

33 of 102 The Handwritten Numbers And Equationsâ&#x20AC;Š

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42 of 102 modelfuture When you write a program that gets the same results as you did by hand, you know you have not made an error. This program, designed to project the future, is more sophisticated than the work I did by hand in the Levinson story, because we can now adjust the probabilities, p and q, and use any destination star we wish. Here is the program, which I call â&#x20AC;&#x153;model future" (modelfuture.c). I have already written programs, bioplanet, model planet and model ocean. Have combined them in one work called star system. They will be presented at the end of this paper.

For now, here is the code in C for model future, and a sample run that shows the calculations in the Levinson story, are accurate.â&#x20AC;Š

43 of 102 #include <stdio.h> #include <math.h> int main (void) { printf("\n"); int N, r; double u, v, y, z; double t,loga, ratio; int n1, n2; char name[15]; float W,fact=1,fact2=1,fact3=1,a,g,rate,T,T1; double x,W2; printf("(p^n1)(q^n2)[W=N!/(n1!)(n2!)]"); printf("\n"); printf("x=e^(c*t)"); printf("\n"); printf("W is the probability of landing on the star in N jumps.\n"); printf("N=n1+n2, n1=number of one light year jumps left,\n"); printf("n2=number of one light year jumps right.\n"); printf("What is 1, the nearest whole number of light years to the star, and\n"); printf("2, what is the star's name?\n"); printf("Enter 1: "); scanf("%i", &r); printf("Enter 2: "); scanf("%s", name); printf("Star name: %s\n", name); printf("Distance: %i\n", r); printf("What is n1? "); scanf("%i", &n1); printf("What is n2? "); scanf("%i", &n2); printf("Since N=n1+n2, N=%i\n", n1+n2); N=n1+n2; printf("What is the probability, p(u), of jumping to the left? "); scanf("%lf", &u); printf("What is the probability, p(v), of jumpint to the left? "); scanf("%lf", &v); printf("What is the probability, q(y), of jumping to the right? "); scanf("%lf", &y); printf("What is the probability, q(z), of jumping to the right? "); scanf("%lf", &z); printf("p=u:v"); printf("\n"); printf("q=y:z"); printf("\n"); for (int i=1; i<=N; i++) { fact = fact*i;

44 of 102 printf("N factorial = %f\n", fact); a=pow(u/v,n1)*pow(y/z,n2); } for (int j=1; j<=n1; j++) { fact2 = fact2*j; printf("n1 factorial = %f\n", fact2); } for (int k=1; k<=n2; k++) { fact3 = fact3*k; printf("n2 factorial = %f\n", fact3); x=2.718*2.718*2.718*2.718*2.718; g=sqrt(x); W=a*fact/(fact2*fact3); printf("W=%f percent\n", W*100); W2=100*W; printf("W=%.2f percent rounded to nearest integral\n", round(W2)); } { printf("What is t in years, the time over which the growth occurs? "); scanf("%lf", &t); loga=log10(round(W*100)); printf("log(W)=%lf\n", loga); ratio=loga/t; printf("loga/t=%lf\n", ratio); rate=ratio/0.4342; //0.4342 = log e// printf("growthrate constant=%lf\n", rate); printf("log 100 = 2, log e = 0.4342, therfore\n"); printf("T=2/[(0.4342)(growthrate)]\n"); T=2/((0.4342)*(rate)); printf("T=%.2f years\n", T); printf("What was the begin year for the period of growth? "); scanf("%f", &T1); printf("Object achieved in %.2f\n", T+T1); } }

45 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): make modelfuture clang -ggdb3 -O0 -std=c99 -Wall -Werror modelfuture.c -lcs50 -lm -o modelfuture jharvard@appliance (~/Dropbox): ./modelfuture (p^n1)(q^n2)[W=N!/(n1!)(n2!)] x=e^(c*t) W is the probability of landing on the star in N jumps.

N=n1+n2, n1=number of one light year jumps left, n2=number of one light year jumps right.

What is 1, the nearest whole number of light years to the star, and 2, what is the star's name?

Enter 1: 4 Enter 2: alphacentauri Star name: alphacentauri Distance: 4 What is n1? 3 What is n2? 7 Since N=n1+n2, N=10 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 2 What is the probability, q(y), of jumping to the right? 1 What is the probability, q(z), of jumping to the right? 2 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 N factorial = 362880.000000 N factorial = 3628800.000000 n1 factorial = 1.000000 n1 factorial = 2.000000 n1 factorial = 6.000000 n2 factorial = 1.000000 W=59062.500000 percent W=59063.00 percent rounded to nearest integral n2 factorial = 2.000000 W=29531.250000 percent W=29531.00 percent rounded to nearest integral n2 factorial = 6.000000 W=9843.750000 percent W=9844.00 percent rounded to nearest integral

46 of 102 n2 factorial = 24.000000 W=2460.937500 percent W=2461.00 percent rounded to nearest integral n2 factorial = 120.000000 W=492.187500 percent W=492.00 percent rounded to nearest integral n2 factorial = 720.000000 W=82.031250 percent W=82.00 percent rounded to nearest integral n2 factorial = 5040.000000 W=11.718750 percent W=12.00 percent rounded to nearest integral What is t in years, the time over which the growth occurs? 40 log(W)=1.079181 loga/t=0.026980 growthrate constant=0.062136 log 100 = 2, log e = 0.4342, therfore T=2/[(0.4342)(growthrate)] T=74.13 years What was the begin year for the period of growth? 1969 Object achieved in 2043.13 jharvard@appliance (~/Dropbox): â&#x20AC;Š

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As climate science is a new science, there are many models for the climate and I learned my climate science at MIT in a free online edX course. One can generate a basic model for climate with nothing more than high school algebra using nothing more than the temperature of the sun, the distance of the earth from the sun, and the earth’s albedo, the percent of light it reflects back into space.

The luminosity of the sun is: 26 ! L0 = 3.9 ×10 J /s

The separation between the earth and the sun is: €

! 1.5 ×1011 m The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:

€

!

S0 =

3.9 ×10 26 = 1,370Watts /meter 2 11 2 4π (1.5 ×10 )

That is the effective energy hitting the earth per second per square meter.

This radiation is equal to the temperature, ! Te, to the fourth power by the steffan-bolzmann constant, sigma ! (σ). ! Te can be called the effective temperature, the temperature entering the earth.

€

€ 2 S ! 0 intercepts the earth disc, € €! πr , and distributes itself over the entire 2 earth surface, ! 4πr , while 30% is reflected back into space due to the

earth’s albedo, a, which is equal to 0.3, so €

S € a) σTe = 0 (1− 4 πr 2 (1− a)S0 4πr 2 !

€

4

€

But, just as the same amount of radiation that enters the system, leaves it, to have radiative equilibrium, the atmosphere radiates back to the surface

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so that the radiation from the atmosphere, ! σTa plus the radiation entering 4 4 the earth, ! σTe is the radiation at the surface of the earth, ! σTs . However, 4 4 ! σTa = σTe

€

€

€

and we have: €

4

4

4

σTs = σTa + σTe = 2σTe

4

1 4

Ts = 2 Te S 4 σTe = 0 (1− a) 4 σ = 5.67 ×10−8 S0 = 1,370 a = 0.3 1,370 (0.7) = 239.75 4 239.75 4 Te = = 4.228 ×10 9 5.67 ×10−8 ! Te = 255Kelvin

So, for the temperature at the surface of the Earth: €

1

! Ts = 2 4 Te = 1.189(255) = 303Kelvin Let’s convert that to degrees centigrade: €

Degrees Centigrade = 303 - 273 = 30 degrees centigrade And, let’s convert that to Fahrenheit: Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit In reality this is warmer than the average annual temperature at the surface of the earth, but, in this model, we only considered radiative heat transfer and not convective heat transfer. In other words, there is cooling due to vaporization of water (the formation of clouds) and due to the condensation of water vapor into rain droplets (precipitation or the formation of rain).

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The incoming radiation from the sun is about 1370 watts per square meter as determined by the energy per second emitted by the sun reduced by the inverse square law at earth orbit. We calculate the total absorbed energy intercepted by the Earth's disc (pi)r^2, its distribution over its surface area 4(pi)r^2 and take into account that about 30% of that is reflected back into space, so the effective radiation hitting the Earth's surface is about 70% of the incoming radiation reduced by four. Radiative energy is equal to temperature to the fourth power by the Stefan-boltzmann constant.

However, the effective incoming radiation is also trapped by greenhouse gases and emitted down towards the surface of the earth (as well as emitted up towards space from this lower atmosphere called the troposphere), the most powerful greenhouse gas being CO2 (Carbon Dioxide) and most abundant and important is water vapour. This doubles the radiation warming the surface of the planet. The atmosphere is predominately Nitrogen gas (N2) and Oxygen gas (O2), about 95 percent.

These gases, however, are not greenhouse gases. The greenhouse gas CO2, though only exists in trace amounts, and water vapour, bring the temperature of the Earth up from minus 18 degrees centigrade (18 below freezing) to an observed average of plus 15 degrees centigrade (15 degrees above freezing). Without these crucial greenhouse gases, the Earth would be frozen. They have this enormous effect on warming the planet even with CO2 existing only at 400 parts per million. It occurs naturally and makes life on Earth possible. However, too much of it and the Earth can be too warm, and we are now seeing amounts beyond the natural levels through anthropogenic sources, that are making the Earth warmer than is favorable for the conditions best for life to be maximally sustainable. We see this increase in CO2 beginning with the industrial era.

The sectors most responsible for the increase are power, industry, and transportation. Looking at records of CO2 amounts we see that it was 315 parts per million in 1958 and rose to 390 parts per million in 2010. It rose above 40s in radiative equilibrium, that is, it loses as much radiation as it receives. Currently we are slightly out of radiative balance, the Earth absorbs about one watt per square meter more than it loses. That means its temperature is not steady, but increasing.

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import comp102x.IO; /** * Here we write a program in java that models the temperature of a planet for a star * of given luminosity. * @author (Ian Beardsley) * @version (Version 01 March 2016) */ public class bioplanet {

public static void bioplanet() { System.out.print("Enter the luminosity of the star in solar luminosities: "); double lum = IO.inputDouble(); System.out.print("Enter the distance of the planet from the star in AU: "); double r=IO.inputDouble(); System.out.print("Enter albedo of the planet (0-1): "); double a=IO.inputDouble(); double R=(1.5E11)*r; double S=(3.9E26)*lum; double b=S/(4*3.141*R*R); double N = (1-a)*b/(4*(5.67E-8)); double root = Math.sqrt(N); double number = Math.sqrt(root); double answer = 1.189*number; IO.outputln("The surface temperature of the planet is: "+answer+ " K"); double C = answer - 273; double F = 1.8*C + 32; IO.outputln("That is: " +C+ " degrees centigrade"); IO.outputln("Which is: " + F + " degrees Fahrenheit");

} } Enter the luminosity of the star in solar luminosities: 1 Enter the distance of the planet from the star in AU: 1 Enter albedo of the planet (0-1): .3 The surface temperature of the planet is: 303.72751882043394 K That is: 30.727518820433943 degrees centigrade Which is: 87.3095338767811 degrees Fahrenheitâ&#x20AC;Š

52 of 102 print("We determine the surface temperature of a planet.") s=float(raw_input("Enter stellar luminosity in solar luminosities: ")) a=float(raw_input("What is planet albedo (0-1)?: ")) au=float(raw_input("What is the distance from star in AU?: ")) r=(1.5)*(10**11)*au l=(3.9)*(10**26)*s b=l/((4.0)*(3.141)*(r**2)) N=((1-a)*b)/(4.0*((5.67)*(10**(-8)))) root=N**(1.0/2.0) number=root**(1.0/2.0) answer=1.189*number print("The surface temperature of the planet is: "+str(answer)+"K") C=answer-273 F=(9.0/5.0)*C + 32 print("That is " +str(C)+"C") print("Which is " +str(F)+"F") joules=3.9*(10**26)*s/1E25 lum=(3.9E26)*s print("luminosity of star in joules per sec: "+str(joules)+"E25") HZ=((lum/(3.9*10**26)))**(1.0/2.0) print("The habitable zone is: "+str(HZ)) flux=b/1370.0 print("Flux at planet is "+str(flux)+" times that at earth") print("That is " +str(b)+ " watts per square meter")â&#x20AC;Š

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54 of 102 #include<stdio.h> #include<math.h> int main(void) { float s, a, l, b, r, AU, N, root, number, answer, C, F; printf("We determine the surface temperature of a planet.\n"); printf("What is the luminosity of the star in solar luminosities? "); scanf("%f", &s); printf("What is the albedo of the planet (0-1)?" ); scanf("%f", &a); printf("What is the distance from the star in AU? "); scanf("%f", &AU); r=1.5E11*AU; l=3.9E26*s; b=l/(4*3.141*r*r); N=(1-a)*b/(4*(5.67E-8)); root=sqrt(N); number=sqrt(root); answer=1.189*(number); printf("The surface temperature of the planet is: %f K\n", answer); C=answer-273; F=(C*1.8)+32; printf("That is %f C, or %f F", C, F); printf("\n"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25); float HZ; HZ=sqrt(joules/3.9E26); printf("The habitable zone of the star in AU is: %f\n", HZ); printf("Flux at planet is %.2f times that at earth.\n", b/3.9E26); printf("That is %f Watts per square meter\n", b); }

55 of 102 Terminal Saved Output 01 (Climate Science by Ian Beardsley) Feb 17, 2016 Last login: Wed Feb 17 06:14:19 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/stelr/stellar ; exit; We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 2 What is the albedo of the planet (0-1)?.5 What is the distance from the star in AU? 1.4142135 The surface temperature of the planet is: 279.223602 K That is 6.223602 C, or 43.202484 F The luminosity of the star in joules per second is: 78.00E25 The habitable zone of the star in AU is: 1.414214 Flux at planet is 1.01 times that at earth.

That is 1379.60 Watts per square meter logout [Process completed] According to my computer simulated program if a star is exactly twice as bright as the sun and reflects 50% of the light received back into space, and is at a distance of square root of 2 AU (AU = average earth-sun separation), then the flux at the planet is the same as that at earth and the planet is exactly in the habitable zone. In this scenario, the average yearly temperature is a cool six degrees celsius.

Let us run this program for other values: Here we choose golden ratio=1.618 for amount of solar luminosities, the golden ratio conjugate =0.618 AU for distance from star, and the same for albedo (percent of light reflected back into space by the planet. It returns an average yearly temperature of approximately fahrenheitcelsius equivalence, -40 degrees F = -40 degrees C: Last login: Wed Feb 17 06:28:41 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/stelr/stellar ; exit; We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 1.618 What is the albedo of the planet (0-1)?0.618 What is the distance from the star in AU? 1.618 The surface temperature of the planet is: 231.462616 K That is -41.537384 C, or -42.767292 F The luminosity of the star in joules per second is: 63.10E25 The habitable zone of the star in AU is: 1.272006 Flux at planet is 0.62 times that at earth.

That is 852.66 Watts per square meter logout [Process completed]

56 of 102 Here is another interesting scenario. It returns yearly average temp, freezing temperature of water: Last login: Wed Feb 17 06:40:34 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/stelr/stellar ; exit; We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 60 What is the albedo of the planet (0-1)?0.605 What is the distance from the star in AU? 7.2 The surface temperature of the planet is: 273.042633 K That is 0.042633 C, or 32.076740 F The luminosity of the star in joules per second is: 2340.00E25 The habitable zone of the star in AU is: 7.745966 Flux at planet is 1.17 times that at earth.

That is 1596.76 Watts per square meter logout [Process completed] Now for one solar luminosity, albedo of golden ratio conjugate, earth-sun separation. It returns 15F. Earth average yearly temperature is 15C: Last login: Wed Feb 17 06:41:43 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/stelr/stellar ; exit; We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 1 What is the albedo of the planet (0-1)?0.6 What is the distance from the star in AU? 1 The surface temperature of the planet is: 264.073395 K That is -8.926605 C, or 15.932111 F The luminosity of the star in joules per second is: 39.00E25 The habitable zone of the star in AU is: 1.000000 Flux at planet is 1.01 times that at earth.

That is 1379.60 Watts per square meter logout [Process completed]

57 of 102 Let us try a planet at mars orbit=1.523 AU with 10 solar luminosities albedo golden ratio conjugate = phi=0.618. It returns about 100 degrees celsius, the boiling temperature of water: Last login: Wed Feb 17 06:46:52 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/stelr/stellar ; exit; We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 10 What is the albedo of the planet (0-1)?0.618 What is the distance from the star in AU? 1.523 The surface temperature of the planet is: 376.162537 K That is 103.162537 C, or 217.692566 F The luminosity of the star in joules per second is: 390.00E25 The habitable zone of the star in AU is: 3.162278 Flux at planet is 4.34 times that at earth.

That is 5947.77 Watts per square meter logout [Process completed]â&#x20AC;Š

58 of 102 print("This program finds the temperature of a planet.")

L0=float(raw_input("Luminosity of the star in solar luminosities? ")) sun=3.9E26 S0=L0*sun r0=float(raw_input("planet distance from star in AU? ")) r=(1.5E11)*r0 S=S0/((4)*(3.141)*(r**2)) a=float(raw_input("What is the albedo of the planet (1-0)?: ")) sigma=5.67E-8 TE=((1-a)*S*(0.25)/(sigma))**(1.0/4.0) delta=float(raw_input("temp dif between two layers in Kelvin: ")) x=delta/TE sTe4=(1-a)*S/4 sTs4=3*(sTe4)-(sTe4)*(2-(1+x)**4)-(sTe4)*(1+((1+x)**4)-(1+2*x)**4) result=(sTs4)/(sigma) answer=(result)**(1.0/4.0) print("planet surface temp is: "+ str(answer)+" K") C=answer-273 F=(1.8)*C+32 print("That is "+str(C)+" C, or "+str(F)+" F") print("flux at planet is "+ str(S)+" watts per square meter")â&#x20AC;Š

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60 of 102 bioplanet by ian beardsley March 03, 2016â&#x20AC;Š

61 of 102 #include<stdio.h> #include<math.h> int main(void) { printf("\n"); printf("\n"); printf("Here we use a single atomospheric layer with no\n"); printf("convection for the planet to be in an equilibrium\n"); printf("state. That is to say, the temperature stays\n"); printf("steady by heat gain and loss with radiative\n"); printf("heat transfer alone.\n"); printf("The habitable zone is calculated using the idea\n"); printf("that the earth is in the habitable zone for a\n"); printf("star like the Sun. That is, if a star is 100\n"); printf("times brighter than the Sun, then the habitable\n"); printf("zone for that star is ten times further from\n"); printf("it than the Earth is from the Sun because ten\n"); printf("squared is 100\n"); printf("\n"); float s, a, l, b, r, AU, N, root, number, answer, C, F; printf("We determine the surface temperature of a planet.\n"); printf("What is the luminosity of the star in solar luminosities? "); scanf("%f", &s); printf("What is the albedo of the planet (0-1)?" ); scanf("%f", &a); printf("What is the distance from the star in AU? "); scanf("%f", &AU); r=1.5E11*AU; l=3.9E26*s; b=l/(4*3.141*r*r); N=(1-a)*b/(4*(5.67E-8)); root=sqrt(N); number=sqrt(root); answer=1.189*(number); printf("\n"); printf("\n"); printf("The surface temperature of the planet is: %f K\n", answer); C=answer-273; F=(C*1.8)+32; printf("That is %f C, or %f F", C, F); printf("\n"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25); float HZ; HZ=sqrt(joules/3.9E26);

62 of 102 printf("The habitable zone of the star in AU is: %f\n", HZ); printf("Flux at planet is %.2f times that at earth.\n", b/1370); printf("That is %.2f Watts per square meter\n", b);printf("\n"); printf("\n"); printf("In this simulation we use a two layer atmospheric model\n"); printf("where equilibrium is maintained by both radiative heat\n"); printf("transfer and convection,\n"); printf("\n"); printf("This program finds the temperature of a planet\n"); float L0,sun,S0,r0,R,S,A,sigma,TE,delta,sTe4,sTs4; float result, answer2, c, f, x; printf("Luminosity of the star in solar luminosities? "); scanf("%f", &L0); printf("Planet distance from the star in AU? "); scanf("%f", &r0); printf("What is the albedo of the planet (0-1)? "); scanf("%f", &A); printf("What is the temp dif between layers in kelvin? "); scanf("%f", &delta); sun=3.9E26; S0=L0*sun; R=(1.5E11)*r0; S=(S0)/((4)*(3.141)*R*R); sigma=5.67E-8; TE=(sqrt(sqrt(((1-A)*S*(0.25))/sigma))); x=delta/TE; sTe4=(1-A)*S/4; sTs4=3*(sTe4)-(sTe4)*(2-(1+x)*(1+x)*(1+x)*(1+x))(sTe4)*(1+(1+x)*(1+x)*(1+x)*(1+x)-(1+2*x)*(1+2*x)*(1+2*x)*(1+2*x)); result=(sTs4)/(sigma); answer2=sqrt((sqrt(result))); printf("\n"); printf("\n"); printf("planet surface temp is: %f K\n", answer2); c=answer2-273; f=(1.8)*c+32; printf("That is %f C, or %f F\n", c, f); printf("flux at planet is %f watts per square meter\n", S); printf("\n"); printf("\n"); }â&#x20AC;Š

63 of 102 Last login: Wed Mar 2 22:55:56 on ttys000 Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/bioplanet ; exit; Here we use a single atmospheric layer with no convection for the planet to be in an equilibrium state. That is to say, the temperature stays steady by heat gain and loss with radiative heat transfer alone.

The habitable zone is calculated using the idea that the earth is in the habitable zone for a star like the Sun. That is, if a star is 100 times brighter than the Sun, then the habitable zone for that star is ten times further from it than the Earth is from the Sun because ten squared is 100 We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 1 What is the albedo of the planet (0-1)?.3 What is the distance from the star in AU? 1 The surface temperature of the planet is: 303.727509 K That is 30.727509 C, or 87.309517 F The luminosity of the star in joules per second is: 39.00E25 The habitable zone of the star in AU is: 1.000000 Flux at planet is 1.01 times that at earth.

That is 1379.60 Watts per square meter In this simulation we use a two layer atmospheric model where equilibrium is maintained by both radiative heat transfer and convection, This program finds the temperature of a planet Luminosity of the star in solar luminosities? 1 Planet distance from the star in AU? 1 What is the albedo of the planet (0-1)? .3 What is the temp dif between layers in kelvin? 30 planet surface temp is: 315.447876 K That is 42.447876 C, or 108.406174 F flux at planet is 1379.603149 watts per square meter

64 of 102 logout [Process completed]â&#x20AC;Š

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modelplanet (bug fixed) March 08, 2016 © 2016 by Ian Beardsley

66 of 102 #include <stdio.h> #include <math.h> int main(void) { printf("\n"); printf("We input the radii of the layers of a planet,...\n"); printf("and their corresponding densities,...\n"); printf("to determine the planet's composition.\n"); printf("Iron Core Density Fe=7.87 g/cm^3\n"); printf("Lithosphere Density Ni = 8.91 g/cm^3\n"); printf("Mantle Density Si=2.33 g/cm^3\n"); printf("Earth Radius = 6,371 km\n"); printf("Earth Mass = 5.972E24 Kg\n"); printf("\n"); float r1=0.00, r2=0.00, r3=0.00, p1=0.00, p2=0.00, p3=0.00; printf("what is r1, the radius of the core in km? "); scanf("%f", &r1); printf("what is p1, its density in g/cm^3? "); scanf("%f", &p1); printf("what is r2, outer edge of layer two in km? "); scanf("%f", &r2); printf("what is p2, density of layer two in g/cm^3? "); scanf("%f", &p2); printf("what is r3, the radius of layer 3 in km? "); scanf("%f", &r3); printf("what is p3, density of layer three in g/cm^3? "); scanf("%f", &p3); printf("\n"); printf("\n"); printf("r1=%.2f, r2=%.2f, r3=%.2f, p1=%.2f, p2=%.2f, p3=%.2f \n", r1,r2,r3,p1,p2,p3); printf("\n"); float R1, v1, m1, M1; { R1=(r1)*(1000.00)*(100.00); v1=(3.141)*(R1)*(R1)*(R1)*(4.00)/(3.00); m1=(p1)*(v1); M1=m1/1000.00; printf("the core has a mass of %.2f E23 Kg\n", M1/1E23); printf("thickness of core is %.2f \n", r1); } float R2, v2, m2, M2; { R2=(r2)*(1000.00)*(100.00); v2=(3.141)*(R2*R2*R2-R1*R1*R1)*(4.00)/(3.00); m2=(p2)*(v2); M2=m2/1000.00; printf("layer two has a mass of %.2f E23 Kg\n", M2/1E23);

67 of 102 printf("layer two thickness is %.2f \n", r2-r1); } float R3, v3, m3, M3; { R3=(r3)*(1000.00)*(100.00); v3=(3.141)*(R3*R3*R3-R2*R2*R2)*(4.00)/(3.00); m3=(p3)*(v3); M3=m3/1000.00; printf("layer three has a mass of %.2f E23 Kg\n", M3/1E23); printf("layer three thickness is %.2f \n", r3-r2); } printf("\n"); printf("\n"); printf("the mass of the planet is %.2f E24 Kg\n", (M1+M2+M3)/1E24); }

68 of 102 I fixed the bug in this program and it models the earth perfectly, exactly returning its mass. Here it is running it on the jharvard emulator: jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): make modelplanet clang -ggdb3 -O0 -std=c99 -Wall -Werror modelplanet.c -lcs50 -lm -o modelplanet jharvard@appliance (~/Dropbox): ./modelplanet We input the radii of the layers of a planet,... and their corresponding densities,... to determine the planet's composition.

Iron Core Density Fe=7.87 g/cm^3 Lithosphere Density Ni = 8.91 g/cm^3 Mantle Density Si=2.33 g/cm^3 Earth Radius = 6,371 km Earth Mass = 5.972E24 Kg what is r1, the radius of the core in km? 500 what is p1, its density in g/cm^3? 7.87 what is r2, outer edge of layer two in km? 5000 what is p2, density of layer two in g/cm^3? 8.91 what is r3, the radius of layer 3 in km? 6371 what is p3, density of layer three in g/cm^3? 2.33

r1=500.00, r2=5000.00, r3=6371.00, p1=7.87, p2=8.91, p3=2.33 the core has a mass of 0.04 E23 Kg thickness of core is 500.00 layer two has a mass of 46.60 E23 Kg layer two thickness is 4500.00 layer three has a mass of 13.04 E23 Kg layer three thickness is 1371.00

the mass of the planet is 5.97 E24 Kg jharvard@appliance (~/Dropbox): â&#x20AC;Š

69 of 102 #include <stdio.h> int main (void) { int option; printf("\n"); printf("The surface area of the earth is 510E6 square km.\n"); printf("About three quarters of that is ocean.\n"); printf("Half the surface area of the earth is receiving sunlight at any given moment.\n"); printf("0.75*510E6/2 = 200E6 square km recieving light from the sun. \n"); printf("There is about one gram of water per cubic cm.\n"); printf("\n"); printf("Is the section of water you are considering on the order of: \n"); printf("1 a waterhole\n"); printf("2 a pond \n"); printf("3 the ocean\n"); scanf("%d", &option); { float area, depth, cubic, density=0.000, mass=0.000; printf("How many square meters of water are warmed? "); scanf("%f", &area); printf("How many meters deep is the water warmed? "); scanf("%f", &depth); cubic=area*depth; density=100*100*100; //grams per cubic meter// mass=(density)*(cubic); if (option==2) { printf("That is %.3f E3 cubic meters of water. \n", cubic/1E3); printf("%.3f cubic meters of water has a mass of about %.3f E6 grams. \n", cubic, mass/1E6); printf("\n"); printf("\n"); } if (option==1) { printf("That is %.3f cubic meters of water.\n", cubic); printf("%.3f cubic meters of water has a mass of about %.3f E3 grams. \n", cubic, mass/1E3); } if (option==3) { printf("That is %.3f E3 cubic meters of water.\n", cubic/1E3); printf("%.3f E3 m^3 of water has a mass of about %.3f E12 g\n", cubic/ 1E3, mass/1E12); printf("\n");

70 of 102 } } printf("\n"); float reduction, incident, energy, watts, square, deep, volume, vol, densiti, matter; float temp, increase, temperature; printf("The specific heat of water is one gram per calorie-degree centigrade.\n"); printf("One calorie is 4.8400 Joules.\n"); printf("The light entering the earth is 1,370 Joules per second per square meter.\n"); printf("That is 1,370 watts per square meter.\n"); printf("By what percent is the light entering reduced by clouds? (0-1) "); scanf("%f", &reduction); incident=reduction*1370; printf("Incident radiation is: %.3f watts per square meter.\n", incident); printf("\n"); printf("The body of water is exposed to the sunlight from 10:00 AM to 2:00 PM.\n"); printf("That is four hours which are 14,400 seconds.\n"); watts=14400*incident; printf("How many square meters of water are to be considered? "); scanf("%f", &square); printf("How deep is the water heated (in meters)? "); scanf("%f", &deep); volume=deep*square; //volume in cubic meters// vol=volume*100*100*100; //volume in cubic centimeters// printf("The volume of water in cubic meters is: %.3f\n", volume); printf("That is %.3f E3 cubic centimeters.\n", vol/1E3); densiti=1.00; //density in grams per cubic cm// matter=densiti*vol; //grams of water// printf("That is %.3f E3 grams of water in %.3f cubic meters of water. \n", matter/1E3, volume); energy=watts*square/4.84; printf("That is %.3f cubic meters heated by %.3f calories\n", volume, energy); printf("What is the intitial temperature of the body of water? "); scanf("%f", &temp); increase=energy/(matter*temp); temperature=increase+temp; printf("The temperature of the body of water has increased; %.3f degrees C.\n", increase); printf("That means the temperature of the body of water is: %.3f degrees C.\n", temperature); }

71 of 102 Running In Utility Terminal Last login: Sat Mar 26 07:26:13 on ttys000 /Users/ianbeardsley/Desktop/Model\ Ocean\ 01/modelocean ; exit; Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/Model\ Ocean\ 01/modelocean ; exit; The surface area of the earth is 510E6 square km.

About three quarters of that is ocean.

Half the surface area of the earth is receiving sunlight at any given moment.

0.75*510E6/2 = 200E6 square km recieving light from the sun.

There is about one gram of water per cubic cm.

Is the section of water you are considering on the order of: 1 a waterhole 2 a pond 3 the ocean 1 How many square meters of water are warmed? 9 How many meters deep is the water warmed? 0.5 That is 4.500 cubic meters of water.

4.500 cubic meters of water has a mass of about 4500.000 E3 grams.

The specific heat of water is one gram per calorie-degree centigrade.

One calorie is 4.8400 Joules.

The light entering the earth is 1,370 Joules per second per square meter.

That is 1,370 watts per square meter.

By what percent is the light entering reduced by clouds? (0-1) 1 Incident radiation is: 1370.000 watts per square meter.

The body of water is exposed to the sunlight from 10:00 AM to 2:00 PM. That is four hours which are 14,400 seconds.

How many square meters of water are to be considered? 9 How deep is the water heated (in meters)? 0.5 The volume of water in cubic meters is: 4.500 That is 4500.000 E3 cubic centimeters.

That is 4500.000 E3 grams of water in 4.500 cubic meters of water.

That is 4.500 cubic meters heated by 36684296.000 calories What is the intitial temperature of the body of water? 68 The temperature of the body of water has increased; 0.120 degrees C. That means the temperature of the body of water is: 68.120 degrees C. logout [Process completed]â&#x20AC;Š

72 of 102 Equilibrium: The Crux of Climate Science Let us say the Earth is cold, absolute zero, then suddenly the sun blinks on. The Earth will receive radiation and start to get warmer. As it gets warmer, it starts to lose some of the heat it receives, and warms slower and slower. There are various mechanisms by which the Earth can lose heat; which ones kick in and by how much they draw heat off the planet, are vary numerous and, vary in a wide spectrum as to the amount of heat, or energy in other words, that they can draw off the planet and, at what rates. We have discussed two mechanisms: radiative heat transfer, and convective heat transfer. An example is the vaporization of the ocean, which is water becoming a gas, or clouds in other words. The heat required to raise its temperature to the point that it vaporizes is one calorie per gram degree centigrade. This represents a loss of heat, or energy, from the sun, that would have gone into warming the planet. As well, when the vaporized water, or what are called clouds, condenses into liquid, this represents another loss of heat-energy that would have gone into warming the planet, for the same reason it takes energy to make a refrigerator cold. This condensation of water vapor to its liquid form, is called precipitation, the formation of water droplets, or what we commonly call rain. The amount of water that is vaporized from the ocean must equal the amount that precipitates, rains back upon the earth, in other words. If these two were not equal, then the oceans would dry up. Back to the warming earth: as it warms, it does so slower and slower as the cooling mechanisms kick in.

Eventually the rate at which the earth warms will slow down to zero. At this point the amount of energy it receives equals the amount of energy it loses and the earth is at a constant temperature. This is called an equilibrium state. For the earth, this should be about 15 degrees centigrade in the annual average temperature. If the earth goes out of equilibrium, that is grows warmer or colder with time, then there can be a great deal of causes for this to happen, and many complex factors must be considered to calculate how long it will take the earth to return to a stable temperature (equilibrium state) and to determine what the temperature of the earth will be when it is back in equilibrium.

From a purely mathematical perspective, equilibrium states can be described by placing a ball in a dish and displacing it to either the left or right: it will roll back and forth until by friction it settles at the bottom of the dish motionless (in an equilibrium state). There can be two types of equilibrium states. One, like we just described, a valley, or two, the reverse: a peak where we have a ball balanced at the apex of a mountain. In this scenario, if I displace the ball to the left or right, it will go out of equilibrium, but never return to equilibrium, like it did in the previous example of a trough: but rather roll down the mountain, never to return.

The earth is currently out of equilibrium, that is, it receives more energy per second than it loses by one watt per square meter. This means the earth is warming. The reason for this is mostly because human activity is putting more CO2 into the atmosphere than should be there, which means the earth retains more heat than it can lose.

Ian Beardsley March 25, 2016â&#x20AC;Š

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â&#x20AC;Š

Climate Modeling with Radiative Heat Transfer And Convection

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Climate Defined: Climate is the statistics of the weather including not just the average weather, but also the statistics of its variability, commonly calculated over periods of a year or more. The progression of seasons is not considered an example of climate variability.

Separating signal from the noise is separating climate from the weather. That is we can say it will be warmer in the summer than in the winter, but we canâ&#x20AC;&#x2122;t forecast the weather for any particular day.

Climate is determined by (1) the energy balance between incoming solar radiation and outgoing infrared radiation. That balance is affected by greenhouse gases, or the composition of the atmosphere, in other words. (2) atmospheric and oceanic convection, the flow or transfer of heat within various substances like, water (the ocean) or gases (the atmosphere). (3) Looking at climate change through geologic time, as a record of climate is embedded in the geological record, the substances that were in the atmosphere during an ice age in other words are recorded in the strata, which can be dated, so we can use this information to model where the climate is going.

Climate Cycles Five billion years ago, when the earth and sun formed, the sun was much cooler than it is today, with an output of about 0.7 of its present output. Yet we know that water existed on the earth in liquid form, when it should have been ice. This is known as the Faint Young Star Paradox; the Earth should have been frozen up to 2.5 billion years ago.

Five hundred and fifty million years ago the Earth went through climate swings, being a snowball and then free of ice. Snowball Earth can be accounted for by positive feedback. Albedo is the percent of incoming radiation that is reflected into space. Snow has a higher albedo, so glaciation, or increased snow, increases the albedo of the earth making it emit more radiation back into a space making it cooler which, in turn, makes more snow, which increases the albedo still more, until you get a runaway icehouse, or Snowball Earth.

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Fifty million years ago the earth reached its thermal maximum (Paleocene-Eocene). It took 20,000 years to develop, and 100,000 years to go away. The earth is cooling from that thermal maximum 50 million years ago.

For the past three million years glacial cycles have been going on with a periodicity of about 20,000 to 100,000 years. They are due to orbital dynamics of the earth. They are the glacial-interglacial cycles caused by eccentricity of the Earth orbit, which is a cycle of one hundred thousand years, the obliquity of the earth or change in tilt which is a cycle of 41 thousand years, and precession or wobble of the earthâ&#x20AC;&#x2122;s spin, which has a cycle of 22 thousand years. The story of the earth has been a story of freezing over for 100,000 years, then briefly warming. We have been in one of these short warm periods for the past 10,000 years, called the Holocene, and it would seem it is responsible for the beginning of civilization 7,000 years ago. The last ice cover was about 18,000 years ago.

Composition Of The Atmosphere Through Time The early Earth Atmosphere was probably predominantly hydrogen and helium (H2, He) but was lost to space. The later atmosphere was due to volcanic emissions, and impact by comets and meteorites (H20, CO2, SO2, CO, S2, Cl2, N2, H2, NH3, CH4). Oxygen came later as a by-product of living organisms. The origin of CO2 was volcanic emissions. It was absorbed by water forming carbonic acid, was deposited in the soil, then underwent reactions to become calcium carbonate:

Lifetime of substances in the atmosphere is given by: Abundance (Gton)/Emissions (Gtons/year) = Lifetime (yr)

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CO2 has a lifetime in the atmosphere of 100 years. CO2 exists in vegitation, soils, oceans, atmosphere and sediments. Lifetime relies in the simple model above relies on abundance and emissions are constant, that they are equilibrium processes.

Structure Of The Atmosphere Divided by vertical gradient of temperature, there are four layers to the Earth Atmosphere: Troposphere at 10-18 kilometers, Stratosphere ending at 50 kilometers, Mesosphere ending at 85 kilometers, then the Thermosphere.

80% of the mass is in the troposphere. In climate science we deal mostly with the troposphere, and a little with the stratosphere.

Heat Distribution Over The Earth, where heat is gained, where heat is lost.

Most of the warming is in the continents, Africa, South America, Canada, Asia. We should see a cooling of the lower stratosphere when we have a warming of the lower troposphere as a part of radiative balance of the planet. Raising the temperature one degree centigrade of a cubic meter of sea water requires 4,000 times more energy input than to raise a cubic meter of atmosphere one degree. Water has a high specific heat, that is it takes one calorie to raise a gram of it one degree centigrade. That is one factor that keeps the Earth from getting too warm. The vast majority of change in the energy climate system has gone into the ocean, mostly into the upper 700 meters.

Water expands when you increase its temperature, like most substances, and the sea rise we are seeing is in part due to that. But most is due to the melting of land ice. Most of the land mass in the Northern Hemisphere, so, in the spring, when there is a lot of plant growth in the Northern Hemisphere, there is a drop in CO2.

Annually, anthropogenic emissions increase CO2 by about nine

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gigatons or 900 terragrams. This is only about 1% of the burden, but it must be remembered that is annual and increases with time.

Precipitation is the product of condensation of water vapor that falls under gravity, like rain, sleet, snow, hail,â&#x20AC;Ś Most of the CO2 is absorbed by the ocean and the decrease its PH, making it more acidic. What are the effects on the corral reefs and plankton?

The temperature is pretty much constant in the tropics throughout the year. Prevalence of ocean in the southern hemisphere keeps that area relatively stable.

During the spring and summer foliage comes out and absorbs CO2, then when leaves fall, and decay, that CO2 is returned to the atmosphere in the Fall. Most cooling is in evaporation of water, especially in the tropics. Radiation is absorbed in the tropics and emitted in the poles. The Ocean and the atmosphere transport absorption in the tropics to the poles, where it is emitted. The tropics absorb more radiation than they emit and the poles emit more radiation than they receive.

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Albedo Albedo is a function of surface reflectivity and atmospheric reflectivity. Atmospheric albedo seems to play the primary role in the overall albedo of a planet. Albedo is the percent of light incident to a surface that is reflected back into space. It has a value ranging from zero to one inclusive. Zero is a black surface absorbing all incident light and one is a white surface reflecting all incident light back into space. Albedo plays a dominant role in the climate of a planet. Let us see if we can find a relationship between composition of a planet and its albedo if not in its distance from the star it orbits and its albedo, even a relationship between its albedo and orbital number, in that albedo could be a function of distance from the star a planet orbits because composition seems to be a function of distance of a planet from the star it orbits. As in the inner planets are solid, or terrestrial, and the outer planets are gas giants. There may be an analogue to the Titius-Bode rule for planetary distribution, but for albedo with respect to planetary number. The inner planets are dominantly CO2, Nitrogen, Oxygen, and water vapor, the outer planets, hydrogen and helium.

1.

2.

3.

4.

5.

6.

7.

8.

9.

Mercury albedo of 0.06 composition 95% CO2 Venus albedo of 0.75 composition clouds of sulfuric acid Earth albedo of 0.30 composition Nitrogen, Oxygen, H20 or water vapor Mars albedo of 0.29 composition CO2 Asteroids Jupiter albedo of 0.53 composition hydrogen and helium Saturn albedo of 0.47 composition hydrogen and helium Uranus albedo of 0.51 composition hydrogen, helium, methane Neptune albedo of 0.41 composition of hydrogen and helium

We see the outer gas giant, which are composed chiefly of hydrogen and helium have albedos around 50%. Earth and Mars, the two planets in the habitable zone, are about the same (30%).

Go to the next page for a graph of albedo to planetary number.

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mercury

0.06

venus

0.75

earth

0.3

mars

0.29

asteroids jupiter

0.52

saturn

0.47

uranus

0.51

neptune

0.41

0.8

0.6

0.4

0.2

0 mercury

earth

asteroids

saturn

neptune

The average for the albedo of the inner planets is: (0.06+0.75+0.3+0.29)/4 = 0.35 This is close to the albedo of the habitable planets Earth and Mars.

The average for the albedo of the outer planets is: (0.52+0.47+0.51+0.41)/4 + 0.4775 ~0.48 This says the outer planets are all close to 0.48~0.5

81 of 102 All this also says, if the planet is solid and habitable it probably has an albedo of around 0.3, otherwise it is an outer gaseous planet and probably has an albedo of around 0.5.

82 of 102 Putting Model Future To Work Ian Beardsley © 2016

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Barnard's Star /ˈbɑːrnərd/ is a very-low-mass red dwarf about six lightyears away from Earth in the constellation of Ophiuchus. It is the fourthclosest known individual star to the Sun (after the three components of the Alpha Centauri system) and the closest star in the Northern Hemisphere.

[16] Despite its proximity, at a dim apparent magnitude of about nine, it is not visible with the unaided eye; however, it is much brighter in the infrared than it is in visible light. (From Wikipedia) Tau Boötis (τ Boo, τ Boötis) is an F-type main-sequence star approximately 51 light-years away[1] in the constellation of Boötes. The system is also a binary star system, with the secondary star being a red dwarf. As of 1999, an extrasolar planet has been confirmed to be orbiting the primary star. (From Wikipedia) Since I have written a computer program that does what I did with alpha centauri, the nearest star (4 light years) I can run the program for other stars. Here I have chosen tau Barnard’s Star six light years away. And Tau Booties 51 light years away.

With my program for predicting the future, I can also weight the probabilities according to whether humanity is doing the right thing or not.

In my first calculation I did before my program, I weighted humanity as just as much awful, as positive: ½ weight against success and ½ weight pulling us towards success. That is all published in my work “A Mystical Whole” which now has the source code for my program and a sample running of it that corroborates my earlier findings. I have now run it for the same star, alpha centauri, with probabilities giving humans more credit to their current state with p1=1/3 against, p2=2/3 for us. Also, I did a run with p1=1/4, p2=3/4. Of course both instances turned up a sooner arrival date for making it to the stars. Those program runs will be available in this program. But first, Tau Bootes.

Ian Beardsley April 04, 2016 12:30 AM

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Usually when you write a computer program, you just input the data and it does everything. But because these calculation uses statistics and not algebra, I have to do guess work and use intuition to formulate the input from the given parameters, like those for these stars. I have made what I call a a “projection diagram” that helps me to figure out the input for a given star. Here are those I made for Barnard’s Star and Tau Bootes:

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Unfortunately I can not run projections for Tau Bootes because my computer can not handle 100 factorial. It returns infinite. I could handle Barnard’s Star just fine, but the best star in my mind was the original I used, Alpha Centauri, for projecting the future. When a star is both the same spectral class as the sun, and the closest star to you, you know it was made for the job! Nonetheless, to hand something like Tau Bootes, I would need a more powerful computer. Here are the runs of the programs for Barnard’s Star, and Tau Bootes:

87 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./modelfuture (p^n1)(q^n2)[W=N!/(n1!)(n2!)] x=e^(c*t) W is the probability of landing on the star in N jumps.

N=n1+n2, n1=number of one light year jumps left, n2=number of one light year jumps right.

What is 1, the nearest whole number of light years to the star, and 2, what is the star's name?

Enter 1: 6 Enter 2: Barnard's Star name: Barnard's Distance: 6 What is n1? 1 What is n2? 7 Since N=n1+n2, N=8 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 2 What is the probability, q(y), of jumping to the right? 1 What is the probability, q(z), of jumping to the right? 2 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 n1 factorial = 1.000000 n2 factorial = 1.000000 W=15750.000000 percent W=15750.00 percent rounded to nearest integral n2 factorial = 2.000000 W=7875.000000 percent W=7875.00 percent rounded to nearest integral n2 factorial = 6.000000 W=2625.000000 percent W=2625.00 percent rounded to nearest integral n2 factorial = 24.000000 W=656.250000 percent W=656.00 percent rounded to nearest integral n2 factorial = 120.000000 W=131.250000 percent W=131.00 percent rounded to nearest integral

88 of 102 n2 factorial = 720.000000 W=21.875000 percent W=22.00 percent rounded to nearest integral n2 factorial = 5040.000000 W=3.125000 percent W=3.00 percent rounded to nearest integral What is t in years, the time over which the growth occurs? 40 log(W)=0.477121 loga/t=0.011928 growthrate constant=0.027471 log 100 = 2, log e = 0.4342, therfore T=2/[(0.4342)(growthrate)] T=167.67 years What was the begin year for the period of growth? 1969 Object achieved in 2136.67 jharvard@appliance (~/Dropbox):

89 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./modelfuture (p^n1)(q^n2)[W=N!/(n1!)(n2!)] x=e^(c*t) W is the probability of landing on the star in N jumps.

N=n1+n2, n1=number of one light year jumps left, n2=number of one light year jumps right.

What is 1, the nearest whole number of light years to the star, and 2, what is the star's name?

Enter 1: 50 Enter 2: taubootes Star name: taubootes Distance: 50 What is n1? 25 What is n2? 75 Since N=n1+n2, N=100 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 2 What is the probability, q(y), of jumping to the right? 1 What is the probability, q(z), of jumping to the right? 2 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 N factorial = 362880.000000 N factorial = 3628800.000000 N factorial = 39916800.000000 N factorial = 479001600.000000 N factorial = 6227020800.000000 N factorial = 87178289152.000000 N factorial = 1307674279936.000000 N factorial = 20922788478976.000000 N factorial = 355687414628352.000000 N factorial = 6402373530419200.000000 N factorial = 121645096004222976.000000 N factorial = 2432902023163674624.000000 N factorial = 51090940837169725440.000000 N factorial = 1124000724806013026304.000000 N factorial = 25852017444594485559296.000000 N factorial = 620448454699064672387072.000000

90 of 102 N factorial = 15511211079246240657965056.000000 N factorial = 403291499589617303175561216.000000 N factorial = 10888870415132690890901946368.000000 N factorial = 304888371623715344945254498304.000000 N factorial = 8841763079319199907069674127360.000000 N factorial = 265252889961724357982831874408448.000000 N factorial = 8222839685527520666638122083155968.000000 N factorial = 263130869936880661332419906660990976.000000 N factorial = 8683318509846655538309012935952826368.000000 N factorial = 295232822996533287161359432338880069632.000000 N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf

91 of 102 N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf N factorial = inf n1 factorial = 1.000000 n1 factorial = 2.000000 n1 factorial = 6.000000 n1 factorial = 24.000000 n1 factorial = 120.000000 n1 factorial = 720.000000 n1 factorial = 5040.000000 n1 factorial = 40320.000000 n1 factorial = 362880.000000 n1 factorial = 3628800.000000 n1 factorial = 39916800.000000 n1 factorial = 479001600.000000 n1 factorial = 6227020800.000000 n1 factorial = 87178289152.000000 n1 factorial = 1307674279936.000000 n1 factorial = 20922788478976.000000 n1 factorial = 355687414628352.000000 n1 factorial = 6402373530419200.000000

92 of 102 n1 factorial = 121645096004222976.000000 n1 factorial = 2432902023163674624.000000 n1 factorial = 51090940837169725440.000000 n1 factorial = 1124000724806013026304.000000 n1 factorial = 25852017444594485559296.000000 n1 factorial = 620448454699064672387072.000000 n1 factorial = 15511211079246240657965056.000000 n2 factorial = 1.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 2.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 6.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 24.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 120.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 720.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 5040.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 40320.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 362880.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 3628800.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 39916800.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 479001600.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 6227020800.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 87178289152.000000

93 of 102 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 1307674279936.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 20922788478976.000000 W=inf percent W=inf percent rounded to nearest integral n2 factorial = 355687414628352.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 6402373530419200.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 121645096004222976.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 2432902023163674624.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 51090940837169725440.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 1124000724806013026304.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 25852017444594485559296.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 620448454699064672387072.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 15511211079246240657965056.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 403291499589617303175561216.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 10888870415132690890901946368.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 304888371623715344945254498304.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 8841763079319199907069674127360.000000 W=-nan percent W=-nan percent rounded to nearest integral

94 of 102 n2 factorial = 265252889961724357982831874408448.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 8222839685527520666638122083155968.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 263130869936880661332419906660990976.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 8683318509846655538309012935952826368.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = 295232822996533287161359432338880069632.000000 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent

95 of 102 W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf

96 of 102 W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral n2 factorial = inf W=-nan percent W=-nan percent rounded to nearest integral What is t in years, the time over which the growth occurs? â&#x20AC;Š

97 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./modelfuture (p^n1)(q^n2)[W=N!/(n1!)(n2!)] x=e^(c*t) W is the probability of landing on the star in N jumps.

N=n1+n2, n1=number of one light year jumps left, n2=number of one light year jumps right.

What is 1, the nearest whole number of light years to the star, and 2, what is the star's name?

Enter 1: 4 Enter 2: alphacentauri Star name: alphacentauri Distance: 4 What is n1? 3 What is n2? 7 Since N=n1+n2, N=10 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 3 What is the probability, q(y), of jumping to the right? 2 What is the probability, q(z), of jumping to the right? 3 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 N factorial = 362880.000000 N factorial = 3628800.000000 n1 factorial = 1.000000 n1 factorial = 2.000000 n1 factorial = 6.000000 n2 factorial = 1.000000 W=131101.968750 percent W=131102.00 percent rounded to nearest integral n2 factorial = 2.000000 W=65550.984375 percent W=65551.00 percent rounded to nearest integral n2 factorial = 6.000000 W=21850.328125 percent W=21850.00 percent rounded to nearest integral n2 factorial = 24.000000 W=5462.582031 percent

98 of 102 W=5463.00 percent rounded to nearest integral n2 factorial = 120.000000 W=1092.516479 percent W=1093.00 percent rounded to nearest integral n2 factorial = 720.000000 W=182.086075 percent W=182.00 percent rounded to nearest integral n2 factorial = 5040.000000 W=26.012295 percent W=26.00 percent rounded to nearest integral What is t in years, the time over which the growth occurs? 40 log(W)=1.414973 loga/t=0.035374 growthrate constant=0.081470 log 100 = 2, log e = 0.4342, therfore T=2/[(0.4342)(growthrate)] T=56.54 years What was the begin year for the period of growth? 1969 Object achieved in 2025.54 jharvard@appliance (~/Dropbox): â&#x20AC;Š

99 of 102 jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./modelfuture (p^n1)(q^n2)[W=N!/(n1!)(n2!)] x=e^(c*t) W is the probability of landing on the star in N jumps.

N=n1+n2, n1=number of one light year jumps left, n2=number of one light year jumps right.

What is 1, the nearest whole number of light years to the star, and 2, what is the star's name?

Enter 1: 4 Enter 2: alphacentauri Star name: alphacentauri Distance: 4 What is n1? 3 What is n2? 7 Since N=n1+n2, N=10 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 4 What is the probability, q(y), of jumping to the right? 3 What is the probability, q(z), of jumping to the right? 4 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 N factorial = 362880.000000 N factorial = 3628800.000000 n1 factorial = 1.000000 n1 factorial = 2.000000 n1 factorial = 6.000000 n2 factorial = 1.000000 W=126142.273438 percent W=126142.00 percent rounded to nearest integral n2 factorial = 2.000000 W=63071.136719 percent W=63071.00 percent rounded to nearest integral n2 factorial = 6.000000 W=21023.712891 percent W=21024.00 percent rounded to nearest integral n2 factorial = 24.000000 W=5255.928223 percent

100 of 102 W=5256.00 percent rounded to nearest integral n2 factorial = 120.000000 W=1051.185547 percent W=1051.00 percent rounded to nearest integral n2 factorial = 720.000000 W=175.197601 percent W=175.00 percent rounded to nearest integral n2 factorial = 5040.000000 W=25.028229 percent W=25.00 percent rounded to nearest integral What is t in years, the time over which the growth occurs? 40 log(W)=1.397940 loga/t=0.034949 growthrate constant=0.080489 log 100 = 2, log e = 0.4342, therfore T=2/[(0.4342)(growthrate)] T=57.23 years What was the begin year for the period of growth? 1969 Object achieved in 2026.23 jharvard@appliance (~/Dropbox):

101 of 102 The Author

102 of 102