It's easy enough. The wheel is spinning very fast at 125 MPH. It does so because the force being applied to it by the road is greater. Therefore, the brake must overcome the force of the road going underneath to lock the wheels. This requires more clamp at faster speed than it does at lower speed. Car's can't ignore the laws of physics just because they roll along.
It's easy enough. The wheel is spinning very fast at 125 MPH. It does so because the force being applied to it by the road is greater. Therefore, the brake must overcome the force of the road going underneath to lock the wheels. This requires more clamp at faster speed than it does at lower speed. Car's can't ignore the laws of physics just because they roll along.
The force of the road is'nt greater, the speed of the road is greater.
If there was a difference in speed of the road and the wheel, a force is generated between the road and the wheel. This force generates an acceleration of the wheel and the wheel turn faster.
no it doesn't work like that, lets do an example:
we have a car with the following parameters.
mass = 1000Kg
max acceleration force = 3000N ( assume constant power simpler calc's )
CdA = 6.24 ( http://en.wikipedia.org/wiki/Automobile_drag_coefficient )
p = 1.204 ( http://en.wikipedia.org/wiki/Density_of_air )
total acceleration force:
Fa = 4000
Fd = 0.5*1.204*6.24*v^2
at max speed:
Fa = Fd
therefore
2.2369 * sqrt( 4000 / ( 0.5*1.204*6.24 ) ) = v ( in MPH )
v = 73 mph
now the load on the tyres:
N = (1000*9.81) / 4 ( assuming even weight distro )
the tyre load is independent of speed, if there was downforce involved there would be a v term in the load equation.
maximum force available to the tyre
u = 0.035 ( coefficient of roll resistance http://en.wikipedia.org/wiki/Rolling_resistance )
F= uN = ( 0.035 * 1000 * 9.81 ) / 4
as the weight of the car does not change the value of F stays the same and therefore the same maximum grip available from the tyre, but if the car had downforce the weight would increase as the car gets faster and the maximum grip would increase at speed.
So the maximum grip and therefore braking capabilities do not change with speed.
{ hope i did that correct }
we have a car with the following parameters.
mass = 1000Kg
max acceleration force = 3000N ( assume constant power simpler calc's )
CdA = 6.24 ( http://en.wikipedia.org/wiki/Automobile_drag_coefficient )
p = 1.204 ( http://en.wikipedia.org/wiki/Density_of_air )
total acceleration force:
Fa = 4000
Fd = 0.5*1.204*6.24*v^2
at max speed:
Fa = Fd
therefore
2.2369 * sqrt( 4000 / ( 0.5*1.204*6.24 ) ) = v ( in MPH )
v = 73 mph
now the load on the tyres:
N = (1000*9.81) / 4 ( assuming even weight distro )
the tyre load is independent of speed, if there was downforce involved there would be a v term in the load equation.
maximum force available to the tyre
u = 0.035 ( coefficient of roll resistance http://en.wikipedia.org/wiki/Rolling_resistance )
F= uN = ( 0.035 * 1000 * 9.81 ) / 4
as the weight of the car does not change the value of F stays the same and therefore the same maximum grip available from the tyre, but if the car had downforce the weight would increase as the car gets faster and the maximum grip would increase at speed.
So the maximum grip and therefore braking capabilities do not change with speed.
{ hope i did that correct }
None of that makes any sense at all. Your understanding of physics is severely flawed. Have you even taken high school physics yet?
Well, I can't pull out all the fancy numbers so I can't really say you're wrong.
My parents say I'm special.
(really I couldn't afford highschool. The government took all our money so I left school)
Forbin it makes perfect sense to me.
And I have tested it in real life with many different cars on different surfaces. The braking force required to lock a wheel is higher at a higher speed and lower at a lower speed.
I've been trying to think of a practical situation that matches the car's situation, but I can't think of anything that's a good fit. The best I could come up with is a freely spinning tire. Spin the tire at 100 MPH, and try to stop it using only your hands. Doesn't really work, even without the road underneath. But, it's easy to stop the wheel spinning at 3 MPH using just your hands. It still doesn't account for the mass of the car or the friction of the road, but I think it still applies.
Road power or smth doesn't exist. I think you're aiming at weight of the wheel (10kg? 20kg?) which rotate very quick and has big kinetic energy. It may induce some ilustration of high forces in mind, but I believe they are really low compared to force from 300kg vertical load on wheel and kinetic energy of 1000kg car pushing it forward. If brakes can go over this grip, rotating 20kg wheel is nothing.
It's EXACTLY ZERO.
A rotating wheel ( with no sliding ) has 2 times the energy of a non-rotating one at the same speed. So if you don't lock, it's completely linear ( in direct proportion ), and when you lock it's still linear. You won't get more at a higher speed.
What is zero?
ps. about F1 HUD in tv, only throttle is shown as it is in real. brakes are always as 100% no matter they brake hard or light.
ps. about F1 HUD in tv, only throttle is shown as it is in real. brakes are always as 100% no matter they brake hard or light.
Zero means no extra torque needed to slow down the wheels at a higher speed.
That has confused me for long.
This man's got it right. The friction forces involved completely outweigh the momentum of the wheels. Likewise, the deceleration of the wheels matches that of the car. Thus, the momentum of the wheels, small as it is, remains proportional to that of the car.
Try this: jack your car up, start the engine, and rev it out in top gear. Step on the clutch, then the brake. Note how quickly the wheels stop. You can try this in LFS too if you manage to roll the car. It doesn't take much force to stop the wheels in an instant.
Those "fancy numbers" are why you go to school, so you can understand this stuff. And last time I checked public high school was paid for through taxes. If you're not going to bother with school, don't bother with trying to argue technical points with people more knowledgeable in technical fields.
I understand it just fine. I have to push harder with my foot when I'm going faster, to get lockup or threshold, in a variety of cars with a variety of tires and braking systems. Nothing will ever change my stance on this, because I trust my foot. I've known the guy for 20 years. He's reliable.
So I can't prove what my feet are telling me with mathematics. Oh my God, my life is ruined. Lighten up just a little bit Forby, it's getting so serious.
Well, this must happen for some reason.
Here is a list of factors that I think will bring more brake input to your test at a higher speed. Most of them are minor things, but better than nothing. ( Firstly let's assume that your feeling is 100% trustable and brake torque can always be directly controled by brake pedal at any time. )
1. Tyre temp changes
This will be taken into consideration only if your tyres gain / lose heat really quickly or you are not pushing hard accelerating / braking / cornering all the time. When a car speed ( even at a straight ) for a long time, the tyres gain more heat, leading to more grip ( if not overcooked or overpressured ).
This factor won't bring less grip in the low speed part of one single slow-down progress. So if you do feel less brake force needed at lower speed in a single progress, forget it.
2. Road unevenness frequency vs tyre / suspension frequency
It's mainly the tyres, I think. But the suspension system may also work in this way. They have fixed freqs, at least no big changes, but road unevenness freq is in direct proportion to speed. So, tyre (susp) freq relatively decreases with speed, which means a smoother progress with less chance of locking.
It's the HIGH freq part only. For the low freq part of road unevenness ( bumps or even jumps ), it's another thing.
3. Centrifugal tyre extention
Tyres are extended at high speed by centrifugal "force", leading to a bigger and more even contact to the road surface, and more grip.
4. Brake temp changes
Just as MadCat360 said in #41, in a good straight the brakes lose much heat, and may be to cold to work. Heat will be regained during braking, leading to lower brake (input) threshold ( or locking ).
To make it ture, you have to get a car with very sensitive brake system which lose heat at very high rates. With normal road cars you're are more likely to overcook the brakes and higher brake input will be needed to make full use of your tyre grip.
5. Engine brakiing torque ( on wheel ) changes
High speed => low gear ratio => low engine braking torque on wheels => higher foot brake force needed
If you're using a 4WD, where brake balance is harder to break, engine braking may be more effective.
Myself have never experienced significant brake threshold changes unless strong aero force involved.
Brake temps too, maybe? High speed braking is after a long straight. The brakes can cool as much as 400-500 degrees on racing cars after a good straight.
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Tyres overheating, watch the .spr !
(40 posts, started )
FGED GREDG RDFGDR GSFDG