A Chair for the Decisive Moment: A Chair for Henri Cartier-Bresson

CHAIR FOR THE DECISIVE MOMENT A CHAIR FOR HENRI CARTIERBRESSON

ZAVEN TITIZIAN MACK PHILLIPS DECEMBER 19, 2015

A Chair for the Decisive Moment

by Zaven Titizian and Mack Phillips

“The [chair] for us is a tool, not a pretty mechanical toy.

In the precise functioning of the mechanical object perhaps there is an unconscious compensation for the anxieties and uncertainties of daily endeavor.

In any case, people think far too much about techniques and not enough about [sitting].�

-Henri Cartier-Bresson, on cameras and seeing The chair is both the photograph and the photographer the composition and the composer In photography, the chair is ubiquitous. The chair has always framed the moment Thus, the chair becomes part of the story The composition, but not the emphasis Nothing is staged, neither is the chair The photographer must blend in, to seem as if they were some permanent fixture Spontaneity and anonymity are paramount Also attention to proportion Attention to detail No other distractions Only the event expressed A precise organization of forms Simply. pragmatically and functionally reinterpreted Which give that event its proper expression Preserving this moment in time The smallest thing can be a great subject; the little human detail can become a leitmotiv The [chair] has a unique relationship with each individual subject There exists a back-and-forth, a conversation between the subject and the [chair] The resulting idiosyncrasies are brought forth in the subject Providing opportunities for the elusive and decisive moment

2

The Var Department, Hyeres, France, 1932

3

Place De L’Europe, Gare Saint Lazare, France, 1932

4

Naples, Italy, 1960

5

In a Train, Romania, 1975

6

Valencia, Spain, 1933

7

PRECEDENTS

The chair’s design and proportions are based on the ubiquituous schoolchair.

Image courtesy of Oliver Apt.

8

Many of the original details of the original chair have been reinterpreted, such as the curvature of the bent legs and the placement of the rivets. Image courtesy of Oliver Apt.

9

The texture and materials used in the chair were inspired by the aesthetics of the Roundish Chair by Naoto Fukasawa.

10

The simplicity of parts and the functionality of the chair was inspired by the Folding Chair by Naoto Fukasawa.

11

DESIGN PROCESS

Chair breakdown options.

12

Sketch of chair brekdown of individual parts.

13

Testing bending techniques for front legs.

14

The tight radius and sharp angle required many thin plies – other thicker materials failed easily.

15

Progression of front leg test pieces.

16

Test peices, laminating front leg and cross piece.

17

Sketch of possible connection techniques.

18

Detailed sketch of connection of seat to frame.

19

Updated breakdown of chair parts.

20

Sketch of fabrication methods for individual chair parts.

21

Laminating back leg.

22

Vacuum-forming chair back.

23

Breakdown of parts, further updated.

24

Final arrangement of chair with connection details.

25

Sanding chair frame.

26

Zaven with assembled chair frame.

27

Sketch of anticipated seating position.

28

Sketch of alternate seating position.

29

First chair test.

30

Second chair test.

31

Finished chair.

32

Back of finished chair.

33

Final assembled chair.

34

Disassembled components of chair.

35

STRUCTURAL ANALYSIS

Orthographic views of finished chair.

PLAN/ELEVATIONS 1:10 PLAN/ELEVATIONS 1:10

36

EXPLODED ISOMETRIC 1:5

Exploded axonometric view of finished chair.

37

testing materials 1/8� baltic birch plywood

$54.10

dowelling

$18.08

cnc time

$10.00

glue, etc.

$41.35

varathane

$25.00

____________________ total

total weight of chair

Final costing of chair.

38

$4.23

$151.87

2.5kg

11

10

6 9

7

12

5

2

8 4

1 3 Simplification of chair for structural analysis purposes.

39

centre of gravity Find centre of gravity: 11

Find volume:

10

6

9

12

5

7

8

2 1

4 3

Laminated Baltic Birch Plywood construction Leg diameter = 31.75mm Seat & back thickness = 9.5mm Density DBIRCH = 680kg/m3 Modulus of Elasticity E = 5900MPa Moment of Inertia of frame members I = 49882.11mm Allowable Bending Stress Fb = 10000KPa Allowable Shear Stress Fv = 1350KPa Allowable Compression Stress Fc = 8300KPa

V1234 = πr2h = π(15.9mm2)(419mm)

= 332780.7mm3 V56 = πr2h = π(15.9mm2)(279mm)

= 221589.1mm3 V78 = πr2h = π(15.9mm2)(267mm)

= 212058.4mm3

V910 = πr2h = π(15.9mm2)(356mm)

= 282744.5mm3 V11

= l•w•h = (191mm)(457mm) (6.35mm) = 829226.5mm3

V12

= l•w•h = (368mm)(432mm) (6.35mm) = 1510272mm3

V TOT

= 4V1234 + 2V56 + 2V78 + 2V910 + V11 + V12 = 4(332780.7mm3) + 2(221589.1mm3) + 2(212058.4mm3) + 2(282744.5mm3) + 829226.5mm3 + 1510272mm3 = 5103405.3mm3

Find individual centres of gravity x1 = 12.7mm x2 = 443.8mm x3 = 21.7mm x4 = 434.7mm x5 = 126.8mm x6 = 33.6mm x7 = 331.5mm x8 = 125mm x9 = 12.7mm x10 = 443.8mm x11 = 228.2mm x12 = 228.2mm 40

y1 = 492mm y2 = 492mm y3 = 12.7mm y4 = 12.7mm y5 = 320mm y6 = 320mm y7 = 116mm y8 = 116mm y9 = 480.3mm y10 = 480.3mm y11 = 511mm y12 = 228.6mm

z1 = 213.4mm z2 = 213.4mm z3 = 209.5mm z4 = 209.5mm z5 = 431.8mm z6 = 431.8mm z7 = 431.8mm z8 = 431.8mm z9 = 619.5mm z10 = 619.5mm z11 = 703.4mm z12 = 477.7mm

centre of gravity Find x ,̄ ȳ, z ̄

27 6m

6m 27

m

x ̄

432mm

m

26

7m

7m

26

m

m

y

= [V1234(x1+ x2 + x3+ x4) +V56(x5 + x6) + V78(x7 + x8)+V910 (x9 + x10) + V11x11+ V12x12] ÷ V TOT = [332780.7mm3(12.7mm+443.8mm+21.7mm+

434.7mm)+221589.1mm3(126.8mm+336mm)+212058.4mm3 (331.5mm+125mm)+282744.5mm3(12.7mm +

443.8mm)+829226.5mm3(228.2mm)+1009498mm3(228.2mm)] ÷ 5103405.3mm3 = 228.4mm

x 368mm

ȳ

457mm

798mm

750mm

z ̄

z

x̄

= [V1234(z1+ z2 + z3+ z4) +V56(z5 + z6) + V78(z7 + z8)+V910 (z9 + z10) + V11z11+ V12z12] ÷ V TOT = [153536.4mm3(213.4mm+213.4mm+209.5mm+209.5mm)+ 102235mm3(431.8mm+431.8mm)+ 97838.2mm3(431.8mm+431.8mm)+ 130451mm3(619.5mm+619.5mm)+ 554272mm3(703.4mm)+1009498mm3(477.7mm)] ÷ 5103405.3mm3 = 470.4mm

191m m

x

= [V1234(y1+ y2 + y3+ y4) +V56(y5 + y6) + V78(y7 + y8)+V910 (y9 + y10) + V11y11+ V12y12] ÷ V TOT = [332780.7mm3(492mm+492mm+12.7mm+12.7mm)+ 221589.1mm3(320mm+320mm)+ 212058.4mm3(116mm+116mm)+ 282744.5mm3(480.3mm+480.3mm)+ 829226.5mm3(511mm)+ 1510272mm3 (228.6mm)] ÷ 5103405.3mm3 = 310.8mm

z y

m

The centre of gravity of the chair is located at the centre back of the chair, directly above the intersection of the planes of

the frame.

z̄

356m m 419m

ȳ

41

dead load, tipping & reactions Fminx

Find mass of chair: mD

= VTOT DBIRCH = 0.0051034053m3 • 680kg/m3 = 3.47kg

This was slightly higher than what we measured (2.5kg), but was not far off.

798mm

PD

Find dead load of chair: PD

z x

x̄

= mDa = (3.47kg)(9.8m/s2) = 34.0N

Find minimum amount of force to tip chair resisted by dead load:

x̄ Fminy

PD ȳ

Fminx

= PD x ÷ ̄ zmax = (34.0N)(228.4mm)÷(798mm) = 9.73N

Fminy

= PD ȳ÷ zmax = (34.0N)(310.8)÷(798mm) = 13.24N

The chair is easiest to tip in the x-direction due to the proximity of the centre of gravity to the back of the chair, but either

way requires very little effort to tip the chair.

798mm

Find the reactions in the legs from the dead load: R1 , R2 = PD (0.5)(ȳ÷ℓ)

= (34.0N)(0.5)(310.8mm÷425.4mm)

= 12.4N

z y

425.4mm

R3 , R4 = PD (0.5)[(ℓ-ȳ)÷ℓ]

= (34.0N)(0.5)[(425.4mm-310.8mm)÷425.4mm]

= 4.6N

R2 42

R1 R3

R4

live load reactions case # 1 Find reactions in chair under live load for upright position: Find live loads:

PL1 PL2 PD PL3 y

332mm

PL1

PL2

166mm

PD

305.4mm

PL3

PL2

= upper legs = 200N

PL3

= lower legs + feet = 130N + 28N = 158N

RL3

= PL3 = 158N

Ma R12

=0 = (PL1ℓL1+ PL2ℓL2+ PDℓD)÷(ℓR12) = ((642N)(332mm)+(200N)(166mm)+(34.0N)(305.4mm)) ÷(465mm) = 552.1N

R34

= PL1+ PL2+ PD - R12 = 642N+200N+34.0N - 543.8N

=332.2N

465mm z a

= head + torso + arms = 90N + 420N + 92N + 40N = 642N

PD = 34.0N

Find reactions

z

150mm

PL1

y

RL3 R34

R12 PL1

PL2

PD

223.2mm 223.2mm

R1 = R2 R1 , R2 = R12÷2 = 552.1N÷2 = 276N

R3 = R4 R3 , R4 = R34÷2 = 332.2N÷2 = 166.1N

z

Because chair is symmetrical in x-direction, reactions are equal on either sides.

Most of the force when sitting upright is carried by the back legs.

x

R1 R3

R2 R4

43

live load reactions case # 2 Find reactions in chair under live load for slouching position: Find live loads: PL1V = 642N PL2V

PL1H

PL1V

PL1H = PL1V tanθ

FfL12 = (PL1V+ PL2V) µLEATHER ON WOOD

= 642N tan30

= (200N + 642N) (0.45) = 371N = 378.9N

PD

PL3V = 158N Ff3 = RL3 µRUBBER ON CONCRETE

= 158N (0.6) = 94.8N PL3H = PL3V tanθ

= 158N tan30

= 91N

RL3 = PL3H

= 158N

PL2

PL3V PL3H

z y

225mm

FfL12

453mm

465mm

PL3H

z a

y

R34

R12 PL1

PD

223.2mm 223.2mm

z x

44

Ma R12

R2 R4

=0 = (PL1VℓL1V+ PL1HℓL1H+ PL2VℓL2V+ PDℓD+ RL3ℓL3- FfL12ℓfL12 - (PL3Vℓ3V)÷(ℓR12) = [(642N)(225mm)+(371N)(798mm)+(34.0N)(305.4mm)+(158N) (450mm)-(378.9N)(453mm)-(200N)(450mm)]÷(465mm) = 559.9N

R34

= PL1V+ PL2V+ PD + PL3V - R12 - RL3 = 642N+200N+158 N+34.0N - 562.5N - 158N

=316.1N

PL2

R1 R3

= 34.0N

Find reactions:

PL2V

305.4mm

PL3V

RL3

PD

PD

450mm

Ff3

PL1V PL1H

798mm

83mm

= 200N

Because chair is symmetrical in x-direction, reactions are equal on either sides.

R1 = R2 R1 , R2 = R12÷2 = 559.9N÷2 = 279.95N

R3 = R4 R3 , R4 = R34÷2 = 316.1N÷2 = 158.05N

live load reactions case # 3 Find reactions in chair under live load for upright position on edge of seat: Find live loads:

PL1

PD

z y

PL1

PD

500mm

305.4mm

PL2 z

RL2

PL2

= lower legs + feet + (upper legs ÷ 2) = 130N + 28N + (200÷2) = 258N

RL2

= PL2 = 258N

Ma R12

=0 = (PL1ℓL1+ PDℓD)÷(ℓR12) = [(742N)(50mm)+(34.0N)(305.4mm)]÷(465mm) = 102.1N

R34

= PL1+ PL2+ PD - R12 - RL2 = 742N+258N+34.0N - 93.8N - 258N = 673.9N

Because chair is symmetrical in x-direction, reactions are equal on either sides.

465mm a

= head + torso + arms + (upper legs ÷ 2) = 90N + 420N + 92N + 40N + (200÷2) = 742N

PD = 34.0N

Find reactions:

PL2

50mm

PL1

y

R34

R12 PL1

R1 = R2 R1 , R2 = R12÷2 = 102.1N÷2 = 51.05N

R3 = R4 R3 , R4 = R34÷2 = 673.9N÷2 = 336.75N

PD

223.2mm 223.2mm

PL2 z x

R1 R3

R2 R4

45

live load tipping cases 1-3

332mm

PL2

Find minimum amount of force to tip chair resisted by live loads:

Fa PL1

Case # 1: Sitting upright, centred 798mm

166mm

PD

150mm

305.4mm

PL3

Fa

240mm

159.6mm

465mm

z y

PD

305.4mm

PL2

465mm z a

y

= (PL1VℓL1V - PL1HℓL1H+ PL2VℓL2V+ PL3VℓL3V+FfL12ℓfL12 + PDℓD)÷ zmax = [(642N)(240mm) - (371N)(798mm)+(200N)(382mm)+(158N) (915mm)+(378.9N)(453mm)+(34.0N)(159.6mm)]÷(798mm) = 320.9N

Fa Case # 3: Sitting upright, at front

PL1

50mm

Fa

b

798mm

PL3V

PD

798mm

915mm

382mm

453mm

PL2V

500mm

Case # 2: Slouching

PL1V PL1H

FfL12

46

b

y

PL3H

= (PL1ℓL1+ PL2ℓL2+ PL3ℓL3+ PDℓD) ÷ zmax = [(642N)(332mm)+(200N)(166mm)+(158N) (150mm)+(34.0N)(305.4mm)] ÷ (798mm) = 351.4N

465mm z

Ff3

Fa

Fa

= (PL1ℓL1 - PL2ℓL2+ PDℓD)÷ zmax = [(742N)(50mm)+(258N)(500mm)+(34.0N)(305.4mm)] ÷(798mm) = 221.2N

S Pin-frame analysis section cut location.

47

pin-frame analysis case #1 Find the load take-down for members in chair for case #1: 101.5mm

*Dead load excluded due to negligible weight

PL1 C

400mm

D

∑MA VB

=0 = PL1 (347.5mm÷624mm) = 742N (347.5mm÷624mm) = 414N

VA

= PL1 (276mm÷624mm) = 742N (276mm÷624mm) = 328N

HA

A

VA

HB

B

245mm

205mm

174mm

400mm

AC

VB

245mm

Find reactions:

CD MC =8.82Nm

PL1

MD =8.82Nm HD =201N

HC =201N

HA=201N

∑V = 0

∑H =0 VC = VA HC = HA = 328N = 201N CD

VC =328N VD =414N AC

VA=328N

175mm

350mm

347.5mm

∑V VD

DB VD =414N HD=201N

= 0

∑H = PL1 - VC HD = 742N - 328N = 414N

DB ∑H HD

VC =328N HC =201N

= 0 = HB = 201N

Find moment in joints C & D: HB =201N MC VB =414N HA =201N VA =328N 48

MD

= VD (205mm) - PL1 (102.5 mm) = 409N (144mm) - 742N (102.5mm) = 8.82Nm = MC = 8.82Nm

=0 = HC = 201N

pin-frame analysis case #2 Find the load take-down for members in chair for case #2:

PL1H

350mm

347.5mm

175mm

101.5mm

PL1V C

D

∑V VA

=0 = 642N

HA

= PL1H (tipping force) = 689N

400mm

AC ∑V VC

=0 = HA -PL1H = 689N - 689N

= 642N = 0N

HA

A

VA

HB

B 205mm

245mm

CD MC =70.6Nm

174mm

VB

CD ∑V VD

PL1V

MD =70.6Nm

= 0

∑H =0 = VC-PL1V HD = HC = 642N 642N

= 0N = 0N

HD =0

HC =0

BC

VC =642N VD =0 AC

= 0

∑H = VA HC

DB PL1H =689N

∑V = 0

∑H =0 VB = VD HB = HD = 0N = 0N

VD =0

(since chair is tipping there will be no reaction forces in point B)

HD =0 VC =642N

Find moment in joints C & D:

HC =0

∑M

=0

∑MC MD

=0 = PL1V (0.1025m) = 70.6Nm

HB =0 VB =0

HA

=689N VA =642N

MC

= MD = 70.6Nm

Under this loading would cause the maximum amount of stress and deflection in member AC.

49

pin-frame analysis case #3 Find the load take-down for members in chair for case #3 175mm

350mm

347.5mm

101.5mm

PL1 C

∑MA VB

=0 = PL1 (522.5mm÷624mm) = 742N (522.5mm÷624mm) = 621N

VA

= PL1 (101.5mm÷624mm) = 742N (101.5mm÷624mm) = 121N

400mm

D

VA

HB

B 205mm

245mm

174mm

VB

CD MC =14.1Nm

MD =14.1Nm HD =74N

HC =74N

VC =121N VD =121N AC

DB VD =121N

PL1

=74N HD VC

=121N HC =74N

245mm

Find reactions

CD ∑V = 0

∑H =0 VD = VC HD = HC = 121N = 74N DB ∑V VB

VA =121N 50

= 0

∑H = PL1 - VD HB = 742N - 121N = 621N

=0 = HD = 74N

Find moment in joints C & D

VB =621N MD

=74N

HA=74N

∑V = 0

∑H =0 VC = VA HC = HA = 121N = 74N

MC HB

=74N HA

400mm

A

HA

VA=121N

AC

= VD (205mm) = 98N (205mm) = 14.1Nm = MC = 14.1Nm

Under this loading would cause the maximum amount of stress and deflection in member DB.

frame analysis case #1

MC =8.8Nm

PL1

MD =8.8Nm HD =202N

HC =201N

VC =328N VD =414N

I = πr4 ÷ 4 = π(15.875mm)4 ÷ 4

= 49882.11mm4

102.5mm PL1

MC =8.8Nm

MD =8.8Nm

P VC =328N

VD =414N

VC =328N

V

0

0 VD =414N Mmax =29.2Nm

M

0

0

-8.8Nm

*Assuming that because the members are glue-laminated birch ply, their modulus of elasticity is half of a solid member of a similar size E =11800MPa ÷ 2 =5900MPa

205mm 102.5mm

Find the load, shear, moment and deflection in critical case (#1) for CD

-8.8Nm

PL1

∆

VC

= 328N

VD

= 414N

Mmax

= (Pℓ ÷ 4) - 8.8Nm = [(742N)(205mm) ÷ 4] - 8.8Nm = 17.4Nm

ƒb

=m÷s = 17400Nmm ÷ 3142.2mm3 = 5537.5kPa

ƒc

=P÷A = 202N ÷ π(15.875mm)2 = 255.1KPa

Fcr = π2E ÷ (ℓ ÷ r) = π2(5900MPa) ÷((205mm) ÷ (7.94mm))2

= 1127.7 MPa ƒv

= 3VD ÷ 2A = 3(414N) ÷ 2(791.7mm2) = 784.4kPa

∆max

= Pℓ3 ÷ 48EI = (742N)(205mm)3 ÷ (48)(5900MPa)(49882.11mm4) = 0.45mm

VC =328N

VD =414N

ƒc ÷ Fc + ƒb ÷ Fb ≤ 1.0

= 255.1kPa ÷ 8300kPa + 5537.5kPa ÷ 10000kPa

= 0.585 Under this loading the shear, moment and deflection within the member would be negligible. This member falls within the acceptable limits for compression and bending stress.

51

frame analysis case #2 Find the actual and allowable axial stress, critical buckling load, and deflection in critical case (#2) for AC

RC =642N

ƒc

= VAp ÷ A = (548N + 548N) ÷ π(15.875mm)2 = 1384.3 KPa

Fcr = π2E ÷ (kℓ ÷ r) = π2(5900MPa) ÷(2(430mm) ÷ (7.94mm))2

= 537.61 MPa 400mm

VA =642N

245mm

RC =548N

438mm

∆max VAp =548N

52

HAp =335N

Pcr

= Fcr x A = (537.61 MPa)(π(0.015875m)2) =425.6 KN

ƒb

=m÷s

s

= 3πr3 ÷ 4 = 3π(15.875mm)3 ÷ 4 = 3142.2mm3

ƒb

=m÷s = 70600Nmm ÷ 3142.2mm3 = 22468kPa

ƒv

= 3HAp ÷ 2A = 3(335N) ÷ 2(791.7mm2) = 397.8kPa

∆max

= Pℓ3 ÷ 3EI = (335N)(438mm)3 ÷ (3)(5900MPa)(49882.11mm4) = 31.9mm

ƒc ÷ Fc + ƒb ÷ Fb ≤ 1.0

= 1384.3kPa ÷ 8300kPa + 22468kPa ÷ 10000kPa

= 2.41

Under this loading the axial stress within the member would be negligible and there would be a small amount of deflection. This member falls outside of the acceptable limits for compression and bending stress, however, the chair is stable under normal use.

frame analysis case #3 Find the actual and allowable axial stress, critical buckling load, and deflection in critical case (#3) for BD VD =121N

ƒc

PL1 =621N

HD=74N

=P÷A = (680N + 599N) ÷ π(15.875mm)2 = 1026 KPa

400mm

Fc = π2E ÷ (ℓ ÷ r) = π2(5900MPa) ÷((436mm) ÷ (15.875mm))2

= 79.9MPa

HB =74N 174mm V

B

=621N

30N

Pcr

= Fc x A = (79.9MPa)(π(0.015875m)2) = 67.87KN

ƒb

=m÷s = 14100Nmm ÷ 3142.2mm3 = 4487.3kPa

ƒv

= 3HBp ÷ 2A = 3(180N) ÷ 2(791.7mm2) = 213.8kPa

∆max

= Pℓ3 ÷ 3EI = (180N)(436mm)3 ÷ (3)(5900MPa)(49882.11mm4) = 17.3mm

111N

ƒc ÷ Fc + ƒb ÷ Fb ≤ 1.0

= 1026kPa ÷ 8300kPa + 4487.3kPa ÷ 10000kPa

= 0.57

680N 421mm

∆max

Under this loading the axial stress within the member would be negligible and there would be a small amount of deflection. This member falls within the acceptable limits for compression and bending stress.

HBp =180N

VBp =599N

53

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54

55

56

57

58

59

60

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IMAGE CREDITS

Place De L’Europe, Gare Saint Lazare, France, 1932 Cartier-Bresson, Henri. Place De L’Europe. Gare Saint Lazare. 1932. FRANCE, Paris. Magnum Photos. Web. 19 Dec. 2015. <http://www.magnumphotos.com/Asset/2S5RYDI9CNRQ.html>.

The Var Department, Hyeres, France, 1932 Cartier-Bresson, Henri. The Var Department. Hyeres. 1932. FRANCE, Paris. Magnum Photos. Web. 19 Dec. 2015. <http://www.magnumphotos.com/Asset/-2S5RYDZCKY50. html>.

Valencia, Spain, 1933 Cartier-Bresson, Henri. Valencia. 1933. SPAIN, Paris. Magnum Photos. Web. 19 Dec. 2015. <http://www.magnumphotos.com/Asset/-2S5RYDI0Q9GT.html>.

In a Train, Romania, 1975 Cartier-Bresson, Henri. In a Train. 1975. ROMANIA, Paris. Magnum Photos. Web. 19 Dec. 2015. <http://www.magnumphotos.com/Asset/-2S5RYDIFH5B6.html>.

Naples, Italy, 1960 Cartier-Bresson, Henri. Naples. 1960. ITALY, Paris. Lomography. Web. 19 Dec. 2015. <http://www.lomography.com/magazine/64790-best-of-the-best-henri-cartier-bresson>.

The Steward, courtesy of Oliver Apt. The Steward. 2010. Edmonton. Oliver Apt. By Max Hurd. Web. 19 Dec. 2015. <http://www.oliverapt.com/team/>.

Roundish Chair by Naoto Fukasawa Fukasawa, Naoto. Roundish Chair. 2015. Mjolk, Japan. Mjolk. Web. 19 Dec. 2015. <http://store.mjolk.ca/index.php?product=Roundish+web&shop=1&c=44.45>.

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