URGENT: anyone good at maths?
25 March 2004, 07:32 PM
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URGENT: anyone good at maths?
right, i've already answered this question but i'm a little dubious at the answer...
10. In a hydraulic brake system on a ProDrive Rally Car, a force of 280N is applied at the brake pedal, which has a leverage of 5:1. The Diameter of the master cylinder is 28mm, whilst the diameter of the 16 wheel cylinder pistons is 40mm. the total efficiency of the system is 88%.
a. what is the Total force exerted by the pistons
b. what is the pedal travel if each piston moves 0.2mm
hmmm i seem to get pedal travel of 52.25cm?????? or is it just a crap question?
thanks
Owain
10. In a hydraulic brake system on a ProDrive Rally Car, a force of 280N is applied at the brake pedal, which has a leverage of 5:1. The Diameter of the master cylinder is 28mm, whilst the diameter of the 16 wheel cylinder pistons is 40mm. the total efficiency of the system is 88%.
a. what is the Total force exerted by the pistons
b. what is the pedal travel if each piston moves 0.2mm
hmmm i seem to get pedal travel of 52.25cm?????? or is it just a crap question?
thanks
Owain
25 March 2004, 09:36 PM
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OWBOW,
The master cylinder travels 6.53mm.
The brake pedal travels 3.26cm.
The distance travelled by the pistons is not affected by the system efficiency.
The force (Newtons) generated at the calipers is calculated thus:
Pressure (pascals 'Pa') = force (Newtons) divided by area(sq mm)
Therefore pressure at master cylinder (28mm dia or 14mm radius) is:
(280 x 5) divided by ((14 x14)x 3.1416) = 1400 divided by 615.75 =2.2736Pa
At the calipers 40mm dia or 20mm radius:
Force (Newton) = pressure (Pa) x area
N =2.2736 x ((20 x20) x 3.1416)
N= 2.2736 x1256.6371
N=2857.15 at 100% efficiency
so is equal to 2514 Newtons per piston at 88% efficiency
The force acting on EACH piston at the calipers is 2514 Newtons.
15085 newtons on each front disc (6 pots)
5028 Newtons on each rear disc (2 pots)
Total braking force 40227 Newtons from a pedal force of 280 Newtons, at 88% efficiency.
I think.
Cheers
MTR
The master cylinder travels 6.53mm.
The brake pedal travels 3.26cm.
The distance travelled by the pistons is not affected by the system efficiency.
The force (Newtons) generated at the calipers is calculated thus:
Pressure (pascals 'Pa') = force (Newtons) divided by area(sq mm)
Therefore pressure at master cylinder (28mm dia or 14mm radius) is:
(280 x 5) divided by ((14 x14)x 3.1416) = 1400 divided by 615.75 =2.2736Pa
At the calipers 40mm dia or 20mm radius:
Force (Newton) = pressure (Pa) x area
N =2.2736 x ((20 x20) x 3.1416)
N= 2.2736 x1256.6371
N=2857.15 at 100% efficiency
so is equal to 2514 Newtons per piston at 88% efficiency
The force acting on EACH piston at the calipers is 2514 Newtons.
15085 newtons on each front disc (6 pots)
5028 Newtons on each rear disc (2 pots)
Total braking force 40227 Newtons from a pedal force of 280 Newtons, at 88% efficiency.
I think.
Cheers
MTR
Last edited by MTR; 25 March 2004 at 09:38 PM. Reason: cannot write formulas on here, so have to write out long hand
25 March 2004, 10:16 PM
#5
Pressure remains constant in the perfect closed system, therefore:
280nx5=1400n, Pressure at master cyl. = force/area = 1400/615.752mm2 = 2.274n/mm2 So this pressure will be present at each slave cyl.
Slave cyl area = 1256.64mm2, so force at one cyl. = pressure x area = 2.274n/mm2 x 1256.64 = 2857.6n So total force at 88% = 2857.6n x 16 x 0.88 = 40235n
(4.1 tonnes!)
Movement of slave cyl = movement of master cyl. (closed system remember!) so pedal movement = 0.2mm x 5 = 1mm It will of course move much more than this due to flexibility, distortion, free play, expansion of flexible brake pipes (even braided) etc.
JohnD
MTR - How did you get to the pedal movement?
280nx5=1400n, Pressure at master cyl. = force/area = 1400/615.752mm2 = 2.274n/mm2 So this pressure will be present at each slave cyl.
Slave cyl area = 1256.64mm2, so force at one cyl. = pressure x area = 2.274n/mm2 x 1256.64 = 2857.6n So total force at 88% = 2857.6n x 16 x 0.88 = 40235n
(4.1 tonnes!)Movement of slave cyl = movement of master cyl. (closed system remember!) so pedal movement = 0.2mm x 5 = 1mm It will of course move much more than this due to flexibility, distortion, free play, expansion of flexible brake pipes (even braided) etc.
JohnD
MTR - How did you get to the pedal movement?
Last edited by JohnD; 25 March 2004 at 10:22 PM. Reason: To say that MTR is as sad as me doing this at this time of night!
25 March 2004, 11:13 PM
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John,
Volumetric displacement of 16 off 40mm diameter caliper pistons moving 0.2mm is:
((20x20)x 3.1416) x 0.2 x 16 = 4021.238 cu mm
Area of cylinder of master cylinder is:
((14 x 14) x 3.11416 = 615.75 sq mm
length of stroke of master cylinder to displace 4021.238 cu mm is:
4021.238 divided by 615.75 = 6.53mm
Pedal lever ratio is 5:1 so
6.53 x5 =32.65mm or 3.26cm pedal travel.
But as you say flexibilty in bulkhead, bushes, mecahnical linkages will play a part.
But its around that figure.
Who's sad
I love it.
Edited to say, my figures would be identical to Johns for the pressure calculations, but I rounded the numberes, instead of working to the full decimal notation.
The last time I did that on here, someone had a pop.
I'm just glad they don't build aircraft.
Cheers
MTR
Volumetric displacement of 16 off 40mm diameter caliper pistons moving 0.2mm is:
((20x20)x 3.1416) x 0.2 x 16 = 4021.238 cu mm
Area of cylinder of master cylinder is:
((14 x 14) x 3.11416 = 615.75 sq mm
length of stroke of master cylinder to displace 4021.238 cu mm is:
4021.238 divided by 615.75 = 6.53mm
Pedal lever ratio is 5:1 so
6.53 x5 =32.65mm or 3.26cm pedal travel.
But as you say flexibilty in bulkhead, bushes, mecahnical linkages will play a part.
But its around that figure.
Who's sad
I love it.Edited to say, my figures would be identical to Johns for the pressure calculations, but I rounded the numberes, instead of working to the full decimal notation.
The last time I did that on here, someone had a pop.
I'm just glad they don't build aircraft.
Cheers
MTR
Last edited by MTR; 25 March 2004 at 11:22 PM. Reason: Just to add comment about rounding up
26 March 2004, 12:25 AM
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ah-ha! saw where i'd gone wrong thanks to you guys, when i was working out the force ratio i multiplied the 20mm by 16 before i squared it! (d'oh!) ...just corrected that and my answers are exactly the same as MTR... thanks f***ing loads guys. excellent!
Owain
excellent!Owain
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26 March 2004, 12:34 AM
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If I understood how the brakes worked I could have contributed 
I got lost when I started reading it as the car having 16 wheels and thinking one cylinder per piston
I got lost when I started reading it as the car having 16 wheels and thinking one cylinder per piston
Last edited by Hanslow; 26 March 2004 at 12:36 AM. Reason: this wine made it all funny!
26 March 2004, 01:34 AM
#9
MTR
I laid in bed and suddenly realised the volume of fluid displaced at the slave cyls = volume displaced at master cyl = 4021.24mm3, and as you say 4021.24/615.75 = 6.53mm. This times 5 = 32.65mm (I made the error of taking the fluid as being a "solid"). All the flex and free movement still applies. All this finally adds up to the fact that I'm double sad for getting back on the PC to post this correction!!
JohnD
I laid in bed and suddenly realised the volume of fluid displaced at the slave cyls = volume displaced at master cyl = 4021.24mm3, and as you say 4021.24/615.75 = 6.53mm. This times 5 = 32.65mm (I made the error of taking the fluid as being a "solid"). All the flex and free movement still applies. All this finally adds up to the fact that I'm double sad for getting back on the PC to post this correction!!
JohnD
26 March 2004, 06:58 AM
#10
Originally Posted by JohnD
MTR
I laid in bed and suddenly realised the volume of fluid displaced at the slave cyls = volume displaced at master cyl = 4021.24mm3, and as you say 4021.24/615.75 = 6.53mm. This times 5 = 32.65mm (I made the error of taking the fluid as being a "solid"). All the flex and free movement still applies. All this finally adds up to the fact that I'm double sad for getting back on the PC to post this correction!!
JohnD
I laid in bed and suddenly realised the volume of fluid displaced at the slave cyls = volume displaced at master cyl = 4021.24mm3, and as you say 4021.24/615.75 = 6.53mm. This times 5 = 32.65mm (I made the error of taking the fluid as being a "solid"). All the flex and free movement still applies. All this finally adds up to the fact that I'm double sad for getting back on the PC to post this correction!!
JohnD
its too early in the morning to understand wot the f**k u lot have just gone on about lolol but it sounds VERY interesting stuff.......
but ...
31 March 2004, 11:31 PM
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I would say if you can't stop in time take to the verge to the left of the car your about to hit.
No good at maths but have avoided a few accidents in this manner.
Tandy
ps. due for a bake upgrade!
No good at maths but have avoided a few accidents in this manner.
Tandy
ps. due for a bake upgrade!
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