While not impossible, it would be a VERY complicated formula.
First problem is the brake drag co-efficient is not the same at all speeds. Your brakes work better at higher speeds. How much better is dependent on all his "what ifs".
One for the physics experts - Stopping distances at different speeds
- Thread starter Baggsy
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While not impossible, it would be a VERY complicated formula.
First problem is the brake drag co-efficient is not the same at all speeds. Your brakes work better at higher speeds. How much better is dependent on all his "what ifs".
Without regard for the confounding variables, it is a simple speed-squared relationship.
10 percent more speed = 1.1 times the original speed, means the distance goes up by 1.1 squared = 1.21 times the original distance, or 21% more.
It scales across the ENTIRE plausible speed range ... as long as you are excluding the confounding variables.
As caboose noted, an attempt to include the confounding variables will quickly make the actual solution impossible. It is better to understand the underlying basic principle (distance to stop depends on speed squared) and then have an understanding - if not the mathematics - of how the confounding variables will influence it.
Most of the confounding variables make matters worse. Slow operator reaction time. Finite time for weight transfer to compress the front suspension and allow full braking power to be reached. Imperfect match between the operator's rate of increase of brake application and the rate of increase of grip on the front due to weight transfer. Imperfect use of the front versus rear brake (imperfect brake balance). Imperfect judgment by the operator of how much grip is available (this is a biggie). Tire, road surface, and brake pad issues. Most of these can be addressed by ABS and it's why I think ABS is a good thing to have on a street bike.
There are some other confounding variables that can reduce the stopping distance at very high speeds ... notably, aerodynamic drag and downforce. With Formula 1 cars, the downforce is a very significant influence on how much braking can be achieved at high speed. With motorcycles, downforce is normally not significant, but the drag is (at high speed). Also, if the aerodynamic center of pressure is higher than the center of gravity, there is a backwards overturning moment that partially overcomes the tendency of a sport bike to lift up the rear wheel and flip over frontwards under braking.
And, of course, if your braking happens to be approaching corner 5 at Mosport (Canadian Tire Motorsports Park), you have not only a very significant uphill to slow you down but the shape of the track imposes a significant downforce that allows even harder braking - the braking zone starts as a downhill flattens out and then goes uphill.
It's best not to rely on the confounding variables to help you out ... just go by the distance varying with speed squared.
10 percent more speed = 1.1 times the original speed, means the distance goes up by 1.1 squared = 1.21 times the original distance, or 21% more.
It scales across the ENTIRE plausible speed range ... as long as you are excluding the confounding variables.
As caboose noted, an attempt to include the confounding variables will quickly make the actual solution impossible. It is better to understand the underlying basic principle (distance to stop depends on speed squared) and then have an understanding - if not the mathematics - of how the confounding variables will influence it.
Most of the confounding variables make matters worse. Slow operator reaction time. Finite time for weight transfer to compress the front suspension and allow full braking power to be reached. Imperfect match between the operator's rate of increase of brake application and the rate of increase of grip on the front due to weight transfer. Imperfect use of the front versus rear brake (imperfect brake balance). Imperfect judgment by the operator of how much grip is available (this is a biggie). Tire, road surface, and brake pad issues. Most of these can be addressed by ABS and it's why I think ABS is a good thing to have on a street bike.
There are some other confounding variables that can reduce the stopping distance at very high speeds ... notably, aerodynamic drag and downforce. With Formula 1 cars, the downforce is a very significant influence on how much braking can be achieved at high speed. With motorcycles, downforce is normally not significant, but the drag is (at high speed). Also, if the aerodynamic center of pressure is higher than the center of gravity, there is a backwards overturning moment that partially overcomes the tendency of a sport bike to lift up the rear wheel and flip over frontwards under braking.
And, of course, if your braking happens to be approaching corner 5 at Mosport (Canadian Tire Motorsports Park), you have not only a very significant uphill to slow you down but the shape of the track imposes a significant downforce that allows even harder braking - the braking zone starts as a downhill flattens out and then goes uphill.
It's best not to rely on the confounding variables to help you out ... just go by the distance varying with speed squared.
Wouldn't this be a mute point since pretty much any bike (or car for that matter) has more braking power than tire grip so the rider would be artificially limiting pad to rotor friction to keep the tire at maximum traction?
Even the crappy brakes on my 650R will lock the tire easily enough. Or are there tires out there grippy enough that it would be impossible?
Like i said, the bike will flip before you reach your theoretical max braking power as calculated with pad/rotor coeffecient of friction and normal force.
Sorry just skimmed the thread and didn't see your post. That's another possibility although my situation I described was slightly different (losing tire traction before losing brakes).
I've never owned a vehicle that had brakes so bad I couldn't lock the tires.
If you grab a fist full of front brakes before you have transferred all the weight to the front you can lock up the front.
Or if there is debris of some sort on the road that decreases traction.. same thing.
In general at what kind of G-Force of deceleration would the rear of a sport bike come off the ground and et to the edge of flipping? Assume the rider isn't trying to do a stoppie but rather trying to maximize braking.
..Tom
On my old Norton 750 Commando with its twin-leading shoe drum brakes I could easily lock the front wheel for a fraction of a second at 60 mph and release before the bike would start to fall over. I'd leave a patch of around 5 to 10 feet so I gather the lockup woulld last a 20th to a tenth of a second. (yes one of the stupid things you do when you are young.)
..Tom
With my quick calculation with lots of assumptions (height of bike cog, height of rider cog, mass of bike, mass of rider, wheelbase) I get 2 g's to have zero weight on rear tire. As soon as it starts to rotate, you will need to quickly reduce your acceleration as the more you rotate, the less force is required to rotate you.
You're correct eventually everything will cancel out and you'll reach equilibrium when the bike is stopped. But your stopping distance greatly depends on the friction between your tires and road and how much brake force can be applied to your rotors to give you maximum braking. I figured the OP was asking how to calculate stopping distance for a given speed which is where that ball park equation would be used.
I'm not trying to complicate things, but to get a better idea of what kind of stopping distances to expect based on just numbers these would be the factors I'd take into consideration. Braking tests are subjective to how good a rider doing it is and road conditions. A pro rider is probably going stop faster than a leisure rider and the track (where most of these bike tests are conducted) have optimal road surfaces and conditions (their asphalt is better than what's on the roads and aren't abused like regular roads).
That seems high. I've heard number like 1.5-1.7 G in MotoGP from 200 mph.
I doubt that the static coefficient of friction of the public roads would support that high a number (more like 1.2 or less? on a decent road), and most people would mess that up.
Would someone be able to calculate the G force from one of the stops on a bike from the 60-0 chart?
The shortest stopped in 104.8 feet and the longest 148.6 feet.
or do you need to know the stopping time for the calculation?
I just use this simple formula...
Which is theoretical and practically pointless, so it fits into this discussion.
I'll tell my broker. Glad I don't trade on any markets that you're on, I might get run down.
I don't know off top of my head but there are three moments that need to balance.
I have a diagram in my photobucket that would help. If I remember I'll dig it up later tonight and elaborate.
Guess you never rode a Harley, they never touch the front brake and in a panic stop, you "lay'er down".
The answer varies according to rider position and the CG, tires, temp., etc.
Did you read ?
I agree that 2.0 G sounds high in reality, but we were talking theory.
Some of the Porsche Sports Cars have a G-Force Displays and 1.2 G constant decelleration is pretty easy to get with the Street tires they come with. I think I have seen slightly higher a few times. I would be surprised if any bikes on the Street could do much better with street legal tires. (Of course sticky track tires would make that much better.)
..Tom
More like 1.1-1.2g
http://www.bbc.co.uk/sport/0/motogp/17486184
They pull more gs in cornering due to the camber effect.
http://www.f1complete.com/content/view/4638/389/
You frequently see MotoGP riders on one wheel under braking.
2gs would be like bench pressing twice your body weight.
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Assuming Constant acceleration (which is doubtful), 1.1 g for 105 feet, 0.8 g for 149 feet. Obviously some of my many assumptions were off in the first calculation.
