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I'm no engineer, but it's my understanding that, as speed rises, the amount of brake force required increases geometrically, so it would take (roughly) twice the force to slow from 80 to 40 than it would to slow from 40 to 0. Therefore braking times increase more than relative to speed. This is why the theoretical maximum throughput of a road (assuming keeping a "safe" distance) occurs at around 17 mph.


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"Show me someone who says that they have never exceeded a speed limit, and I'll show you a liar, or a menace." (Austin Williams - Director, Transport Research Group)

Any views expressed in this post are personal opinions and may not represent the views of Safe Speed

I think you are confusing force and energy. If you had to press harder at high speed then you would have to back off the brake pedal to stop the wheels locking as you slowed down. Assuming you started at a high level of braking of course.

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F = m * a (force = mass x acceleration)

The same force is required to slow from 80 to 40 (in say 4 seconds) as 40 to 0 (in that same 4 second timeframe).

The brake force is proportional to the rate of deceleration, regardless of the actual speed.

The power being dissipated is proportional to the speed. Three times the energy needs to be dumped (as heat through the disks) from 80 to 40 compared to 40 to 0 (assuming no significant drag losses)

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Views expressed are personal opinions and are not necessarily shared by the Safe Speed campaign

The braking force is constant with speed, depending only on the amount of pedal pressure applied. However, as the brake disks are moving at twice the speed relative to the pads at twice the vehicle speed, the amount of energy dissipated by the brakes is twice as much per unit time with the same force. And, as the vehicle takes twice the time to stop from twice the speed, the total energy dissipated is 2 x 2 = 4 times as much - which fits in nicely with the formula: kinetic energy = 0.5 * mass * (speed) squared

But with modern brakes the limiting factor is not the braking force but almost always the friction coefficient between the tyres and the road. Nonetheless, the deceleration is virtually linear with speed, so you'll take twice the time to brake from twice the speed. (which equates to 4 times the distance)

I don't see what this has got to do with the throughput of the road, though.

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Only when ideology, prejudice and dogma are set aside does the truth emerge - Kepler

Well it looks like everyone hs beaten me to it. I have little to add apart from asking what you mean by "braking force"? Everyone has assumed it means the retardation force acting backwards on the car, through it's tyre contact patches. Is that what you mean? (I just wondered if you perhaps meant the force that the driver must apply to the brake pedal)!

Anyway, the other way to look at braking force is that it's the force you feel "pulling" you towards the front of the car. (That being equal and opposite to the force acting backwards on the car (and everything in it) through it's tyre contact patches.

They're the same thing, just expressed differently.

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Only when ideology, prejudice and dogma are set aside does the truth emerge - Kepler