Mathematics on undocking the soyuz from iss rev2

Mathematics on undocking the Soyuz from ISS

The success of security at entry into the Earth's atmosphere of a vehicle from the Space International Station (ISS), to enter the altitude of 120kms from the atmosphere depends on the entry angle (θ) which is defined by the direction of the velocity vector and the horizontal at the entry point into the atmosphere.

It is with the entry of the vehicle in the altitude of 120kms in the atmosphere, it begins to produce its slowdown due to the increasing density of the air, and so are obtained braking friction that allows a landing with an appropriate speed.

The International Station (ISS) rotates in the direction of Earth's rotation to a 7.652m / s.

Any vehicle from the international station and want to return to earth as a consequence of the law of gravitation has to reduce its speed allowing you to enter a downward path in the form of ellipse, going intercept atmospheric layer 120kms with a speed of the order of 7696 m/s and at an angle θ of well exact entry, according to the value of said speed reduction.

Decoupling scheme and trajectories Fig 1

A - Let's calculate the descent of the Soyuz ellipse until the entry into the Earth's atmosphere.

In the schematic figure above, fig1, there is shown schematically stages of ISS decoupling vehicle, although not exactly correspond to reality. According to an accurate description of the scheme is best described as follows: 1- It starts decoupling impulse springs, pushing the vehicle in the direction opposite to the movement of the ISS, moving it slowly this.

2 About 9 minutes after separation a small boost (ΔV1 = 0.5 m/s) by retro-rocket in the opposite direction to the movement of the vehicle decreases its speed by placing it in an intermediate elliptical trajectory, just below ISS. 3. The intermediate trend lasts about 2-3 hours (two revolutions) and serves to set the new final push point for retro-rocket (ΔV2 = 120m/s) in the opposite direction to the movement of the vehicle and the previously calculated exact coordinates, leading to the vehicle entering the final trajectory to cross the atmosphere at 120kms high, and with an angle θ of appropriate input and to land in the default location.

The value of the pulse (ΔV2) for reducing the speed is critical, because: a) If it is above a calculated threshold, the vehicle plunges more than necessary and enters the layer 120kms with high angle θ with disastrous consequences over-temperature friction and slowdowns to endanger the crew. b) If the impulse is below the calculated limit, the disaster is, the vehicle is bounced and can lost in space.

Trajectory calculations: The ISS rotates in the direction of Earth's rotation with a speed of: VISS = (μ / (Ri)) ^0,5 = 7.652m/s Where: μ constant of gravitation the Earth = 3,97E + 14 Ri movement radius of the ISS (Earth's radius + altitude ISS) = 6,380,000 + 400,000 = Re input beam (Earth's radius + 120,000) = 6.500.000m

6.780.000m

To calculate the trajectory, transfer ellipse, we have to calculate the following ellipse parameters: - Perihelion Altitude (Ap) and perihelion distance (Rp = 6,380 + Ap) = simulation value - Ray desorbitização (Ri = 6,380,000 + 400,000) = 6.780.000m - Semi-major axis of the ellipse A = (R + Rp) / 2 m - Eccentricity of the ellipse (e = 1-Rp / a) - Desorbitização speed Vi = (μ * (2 / Ri - 1 / a) ^ 0.5) m / s - Speed perihelion Vp = μ * (2 / R p - 1 / a)) ^ 0.5 m / s - The elliptical motion period T = 6.28 * ((a ^ 3 / μ) ^ 0.5 / 3600 hours - ΔV2 = VISS - Vi m / s - Angle traversed cosα = (a * (1-e ^ 2) / (Re * e) - 1 / e) radians - Α angle α traveled degrees - Angular momentum of the body L = Ri * Vi N.m - Approximate time t = 60 * (T / 2) * (α) / 180 minutes - Vα = L / Re m/s - Input speed Ve = 2 * μ * (1 / Re - 1 / (2 * a))) ^ 0.5 m / s - Cosθ = Vα / Ve radians - Entry angle θ. degrees

In these formulas it was constructed in Excel a simulator which simulate (green cell) the perihelium radius in order to obtain ΔV2, to obtain an acceptable angle θ input.

altitude ISS AISS VISS = (µ/(Ri))^,5 radius of the incoming atmospheric layer perihelion altitude Ap constant gravitation earth μ1 Rp=6380000+AP Ri=6380000+AISS a= (Rp+Ri)/2 e=(1-Rp/a) Vi=(µ*(2/Ri-1/a))^0,5 Vp=(µ*(2/Rp-1/a))^0,5 T=6,28*((a^3/µ)^0,5/3600 Rp=a*(1-e^2)/(1-e*cos(180)) ΔV2= Viss-Vi Re=a*(1-e^2)/(1-e*cosα) cosα=(a*(1-e^2)/(Re*e)-1/e) α angle α traveled degrees graus acos Angular momento body L=[μ1.a.(1-e²)]½ Angular momento body L=Ri*Vi Approximate time t = 60*(T/2)*(α)/180 Vα = L/Re Ve=(2*µ*(1/Re-1/(2*a)))^0,5 Cosθ= Vα/V θ entry angle =

400 000 7 652 6 500 000 -10 000 3,97E+14 6 370 000 6 780 000 6 575 000 0,0312 7 532 8 017 1,48 6 370 000 120 6 500 301 0,3385 109,75 51 066 011 406 51 066 011 406 27,00 7 856 7 859 0,9996 1,66

m m/s m m m m m m/s m/s Hours m m/s m degrees N*m N*m mi m/s m/s radians degrees

Summary simulations:

Ap

ΔV2

θ

m

m/s

graus

0 -5000 -10000 -20000 -30000

117 119 120 123 126

1,60 1,63 1,66 1,73 1,79

B- Explained Descent from the altitude of 120kms. (descent module of the Soyuz).

The crossing of the atmosphere after entry into the 120kms of altitude, can be made:

a) In ballistic trajectory, that is, only with the descent module oriented with the protective heat shield perpendicular to the velocity vector and a small rotation about the main axis towards the moment of inertia thus obtained help decrease in its stability. b) piloted path, with the heat shield inclined at an angle (AP) for the velocity vector (to see below) which allows its planing while crossing the atmosphere and thus reduce the effects of deceleration of the crew. The control of the steering angle (PA) is done by changing the module slope (see fig5).

The two essential aspects to take into account the effects of the slowdown are: 1- The temperature (can reach to 2100ยบC) friction in the atmosphere in the protective module shell. It is not the subject of study in this article.

2. The value of the module slowdown, which has physiological implications for the crew. A slowdown in the range of 4-5 the severity value is acceptable for several minutes. 9 times a deceleration value of gravity is only permissible for a few seconds.

The movement during the course of the study in the atmosphere can be done by analytical mathematical equations (by differential equations), but is complicated and outside the scope of this article.

Let us follow another method (approximate) by numerical integration using basic physical formulas being more simple and plain, and gives a very approximate solution.

The description of the descent can be seen in "YouTube" in the "LINK". .

https://youtu.be/-l7MM9yoxII In the figure one can see the assembly of the three modules Soyus.

The descent vehicle and that will be the subject of this study is shown in the following figures (descent module of the Soyuz) is what makes the entry into the Earth's atmosphere.

Figure2 Characteristics: About 2,7m diameter at the base Height: about 2.1 m Weight: 2877 kg

Figure 3

Figure 4

Although the three modules that are decoupled from the ISS, only the descent module (fig3 and 4) enters and travels in a controlled manner into the atmosphere. The other two are disengaged before entry into the atmosphere of the descent module and will be destroyed by thermal effects while they go through the atmosphere.

C- Assumptions used in calculating the trajectory by numerical integration after the entry 120kms above sea level: 1. To conduct a numerical integration, it is considered a system of fixed axes (X and Y), where the various forces involved will be projected during the descent.

The vertical entry point into the atmosphere 120kms defines the Y-axis (positive towards the center of the earth. The X-axis is perpendicular to the Y axis, and positive in the direction movement (see Figure 5 below). The origin the axis is vertical to the ground entry point. The system of axes is integral with the movement of the earth's rotation.

2 is neglected since the earth's sphericity length between the module entry point into the atmosphere and fall to the ground is relatively small.

3. Let us consider the trajectory situated in a plane defined by the vector which connects the center of the earth to the point of entry to 120kms altitude and velocity vector at the entrance. The analysis is thus made in this plan.

4. The entrance angle θ is defined as the angle between the velocity vector V and the axis X. The steering angle (PA) is defined as the angle between the velocity vector V and the main axis of the module .

5- During its penetration into the atmosphere the module is subjected to four forces applied at its center of gravity. It should be noted that the module's center of gravity position is low, allowing greater stability in the module's descent.

6- Schema distribution of instantaneous forces on the module

Figura5

6.1- The drag force is determined by the physics of the formula FD = 0,5.Cd.ρ.A.V² oriented vector in the opposite direction to the velocity vector.

6.2- The lifting force FL = 0,5.Cl.ρ.A.V², vector perpendicular to the velocity vector 6.3- The force of gravity is determined by the physics formula FG = mg with the vector directed in the positive direction of the shaft axis Y. 6.4-The centrifugal force caused by the curvature of the velocity vector is determined by the physics formula.

FC = (m.v²) / (Rt + H). The vector is variable in time (curvature of the path) at the downward angle (θ) and perpendicular to the vector V. Where: ρ - air density (kg / m3), varying with the altitude (H, m) A - area of the heat shield collision section in the direction of movement (m2) V - Module Speed (m / s) variable.

M - module weight (kg), constant. g - Gravity, varying with the altitude. (m / s2) Rt - radius of the earth (m) H - Height above sea level (m) variable.

Cd and Cl - are respectively the experimental aerodynamic lift and drag coefficients, depending on the geometrical shape of the module and the steering angle (PA). The steering angle (PA) is the angle formed by the velocity vector and module axis (see figure above).

It was considered Cl and Cd values approximate and obtained the literature, as detailed in the table below.

As the four forces are not constant during the penetration trajectory, the motion is variable (though not uniformly variable).

D re-entry schemes in the atmosphere.

There are several possible paths. Let's just consider some taking into account: a) The ballistic entry is made with the steering angle AP = 0 b) piloted entry is made so that the module travels a trajectory with some plane. Due to its geometric shape with penetration angle (θ) and pilot (AP) suitable module can glide (due to the lifting component) and so reduce the forces of drag and consequently the physical strain on the crew and thermal effects.

Assumptions of the method by numerical integration: It divides the route into successive atmospheric layers traversed in the range for example of 1 second of time.

Thus, for every second that the module runs have a small gap height (∆H) and a length (ΔL), and

that range is subject to four forces, the drag force (FA), the lifting force (FL), the force of gravity (FG = mg) and the centrifugal force (FC) of curvature of the velocity vector.

In this short break we can consider that, ρ, V, and g, are practically constant, although the altitude with values given by the following formulas of physics: ρ (H) = ρº.EXP (-28,9.g.H / (R.T)) R the gas constant and T is the absolute temperature. (° Kelvin) g (H) = μ / (Rt + H) ² being, Rt the radius of the earth and μ the universal constant of the earth.

H range, a movement which is characterized by the action of a substantially constant force resulting from the vectorial composition of the forces (FL + FD + FG + FC).We therefore the H range, the resultant force then produces a uniformly retarded motion and we can apply the formulas of the already known physics.In If we project the forces in a Cartesian axis system as already indicated, we can write the following motion formulas.

FX = -FD.cos(θ)+FL.sen(θ) +FC.sen θ FY = -FD.sen(θ)+ M.g – FL.cos(θ)- FC.cos θ VX = FX.t/M VY = FY.t/M θ = Atang (VY/VX) aX = VX/t aY = VY/t H = VY.t + 0,5.aY.t² L = VX.t + 0,5.aX.t²

Force axis X Force axis Y inside H inside H X aceleration in H Y aceleration in H Y covered space in H X covered space in H

H successive intervals until you get the full range H, which is done in the simulator in Excel sheet from these formulas is the numerical integration

E- results of the simulator. Cells green let you simulate other data.

Data: Module weight: 2.877kgs Area of 5.72 m2 heat shield Input speed 7.835m/s Entry angle (θ): First parachute 24 m2 to automated 25.000m by the simulator large parachute 1,000 m2 to automated 3,000m by the simulator.

AP steering angle automated piloted by the simulator in the entry.

can be simulated can be simulated can be simulated can be simulated

E.1 Consider a ballistic entry (AP = 0).

Simulator result for θ input angles: Results:

angulo graus θ 2,50 2,00 1,66 1,00

tempo mi 16,45 17,13 17,77 19,55

longit kms 1872 2219 2527 3466

desaceleraç vezes g -9,87 -9,13 -8,75 -8,27

For angles greater than 2.5 degrees, the simulator indicates values for the worst downturns.

Therefore they shall not be considered.

The ballistic entry, there is the possibility of a safe entry only corridor of 1.0 ° to 2,0º. An angle θ entry above 2 degrees, the deceleration reaches above 9g values, unaffordable for humans. Below 1 degree, the module is thrown out of the earth. The fall time approximately ranges between 16 and 20 minutes and the decrease in length is between about 2,000 to 5,500 miles.

As can be seen in the simulator slowing up the -4G up to -9,5g, occurs at about 35 km altitude, and only for about 60 seconds (see Excel calculation) The speed of arrival near surface 6m / s (22Km / h). Retro-rockets are driven to further reduce the landing speed.

In the following figure the graphical ballistic scheme for a 1,66º entry angle. altura

angulo

acele

pilotagem

veloci

140 120 100 80 60 40 20 0 1

71

141 211 281 351 421 491 561 631 701 771 841 911 981 1051 1121 1191 1261

-20

E.2 entry piloted. The simulator calculates the steering angles (AP) so that the deceleration values do not exceed the downturns of -4G. The way to achieve a lobby with comfortable deceleration is accomplished using the input method with planing with a suitable steering angle (PA) module.

The module has a steering system (computer) that selects a steering angle (AP) which is sensitive to deceleration (in practice the module obtained by accelerometers) to obtain a gliding not allowing high decelerations.

The steering angle is achieved by short pulses of small rockets the module fuselage.

The simulator obtains the following values:

angulo θ 4,00 3,00 2,00 1,66 1,50

tempo mi 21,55 22,77 24,38 25,15 25,53

longit kms 2691 3300 4259 4739 5015

aceleraç vezes g -4,26 -3,63 -3,62 ~3,63 -3,63

With angle θ above 4 degrees, the simulator indicates deceleration above -4G. Below 1.5 ° degree the module is likely to escape into space.

The graph below represents the glide path scheme with input angle θ = 1.66 degree

altura

angulo

acele

pilotagem

veloci

140 120 100 80 60 40 20 0 -20

1

96

191

286 381 476 571 666 761 856 951 1046 1141 1236 1331 1426 1521 1616 1711

The piloted entry, and allows θ, more open angles, it also allows more security and comfort for the crew.

Gabriel Leite, 8 October 2015

The simulators can be sent, requesting by mail: gabrielleite100@gmail.com