Brakes
07-30-2005, 01:30 PM
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Brakes
I was driving home today quite quickly on a decent stretch of single carriageway NSL, and imagined what it would be like to be cruising at 100mph on a downhill bend (like the one in GT4's Deep Forest before the uphill section, PS2 fans), and meet parked cars waiting on the car in front to make a right-hand turning.
I then imagined how hard my brakes would have to work in such a situation, and whether they could take me from 100mph-0mph in the space of (for talking's sake), say 30 metres. In this purely fictional imagination, my brakes did the job perfectly, dispite me planning on changing the discs once I'm back from my holidays.
An imagination is a wonderful thing.
I then imagined how hard my brakes would have to work in such a situation, and whether they could take me from 100mph-0mph in the space of (for talking's sake), say 30 metres. In this purely fictional imagination, my brakes did the job perfectly, dispite me planning on changing the discs once I'm back from my holidays.
An imagination is a wonderful thing.
07-31-2005, 07:01 AM
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As we all know the equations of motion say V^2=U^2+2AS for a body moving ins a straight line under constant acceleration.
V = final velocity = 0 so in our case V^2=0
U = initial velocity = 100mph = 44.7 mtrs per sec so U^2 = 1998
S = distance covered during the change in speed = 30 mtrs in this case
So solving for A, the acceleration (in this case deceleration so a negative figure) we have A = -1998/(2 x 30)
Gives us an answer of minus (indicating deceleration) 33 metres per second per second.
g the acceleration due to gravity is just under 10 metres per second per second - so that is over 3g
Seems a little high to me! 1g maybe. Road cars cannot achieve much above 1g as they are dependent on the friction between tyre and road - which is in turn dependent on the downforce on the tyres - which for road cars is approximately the mass of the car. At 1g the braking force is the same as the cars mass implying a coefficient of friction for rubber of about 1.
F1 cars can exceed 1g as they have downforce to add load to the wheels and so increase the available braking force - though even they could not sustain over 1g all the way to rest as the downforce vanishes as the car slows. Effectively they can exert a downforce on the tyre of several times the bodies mass allowing the tyre rubber to apply a force of several times the cars mass in both the fore and aft or sideways directions. That is why they stick so well. When they brake this braking force of several times the cars actual mass is applied to stopping the cars mass - and so the car can decelerate at several g, at least while the aerodynamics keep working.
At a more reasonable 1g the stopping distance would have been approx 100mtrs.
V = final velocity = 0 so in our case V^2=0
U = initial velocity = 100mph = 44.7 mtrs per sec so U^2 = 1998
S = distance covered during the change in speed = 30 mtrs in this case
So solving for A, the acceleration (in this case deceleration so a negative figure) we have A = -1998/(2 x 30)
Gives us an answer of minus (indicating deceleration) 33 metres per second per second.
g the acceleration due to gravity is just under 10 metres per second per second - so that is over 3g
Seems a little high to me! 1g maybe. Road cars cannot achieve much above 1g as they are dependent on the friction between tyre and road - which is in turn dependent on the downforce on the tyres - which for road cars is approximately the mass of the car. At 1g the braking force is the same as the cars mass implying a coefficient of friction for rubber of about 1.
F1 cars can exceed 1g as they have downforce to add load to the wheels and so increase the available braking force - though even they could not sustain over 1g all the way to rest as the downforce vanishes as the car slows. Effectively they can exert a downforce on the tyre of several times the bodies mass allowing the tyre rubber to apply a force of several times the cars mass in both the fore and aft or sideways directions. That is why they stick so well. When they brake this braking force of several times the cars actual mass is applied to stopping the cars mass - and so the car can decelerate at several g, at least while the aerodynamics keep working.
At a more reasonable 1g the stopping distance would have been approx 100mtrs.
07-31-2005, 08:24 AM
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Originally Posted by GTFCTIM,Jul 31 2005, 04:07 PM
From what I seem to understand from your post, you seem to be saying its possible to stop the S from 100mph in 30 metres?
unless you involve a wall, and maybe 10metres is enough..
07-31-2005, 08:54 AM
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If the wall was solid enough the rear end can stop in about 4 metres - a car's length. The front stops in less distance - about 4 metres less.
Theory apart, the S still has good brakes.
Highway code stopping distances are based on 0.7g I seem to recall. We can beat that!
Makes you think though - 100mtrs at that sort of speed and that is only once you have got your foot to move.
Theory apart, the S still has good brakes.
Highway code stopping distances are based on 0.7g I seem to recall. We can beat that!
Makes you think though - 100mtrs at that sort of speed and that is only once you have got your foot to move.
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07-31-2005, 10:46 AM
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Originally Posted by baptistsan,Jul 31 2005, 06:07 PM
Ah, but then we digress into reaction times.
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