mustang1

I'm just gonna chuck some very rough figures out there, let me know what you think.

I always figured cars have more braking power than bikes, and I guess this is due to more of the car's rubber in contact with the tarmac. So here are my figures:

Car weighs 2000kg (that's 2 metric tons, about 3700 lbs) including some passengers and luggage, has 225mm wide tires at each corner, and about 2cm per tire cross-section is in contact with the ground. So 225mm x 4 tires = 100cm. Times that by 2cm = 200sq cm of total rubber contact with tarmac. That makes 200sq cm / 2 tons = 100sq cm per ton of rubber contact.

Take one racing bike, with rider, total weight = 100kg (that's about 210 lbs, 0.1 metric tons). Tires is 700x25mm. Because of roundness of tire, not all 25mm touches tarmac, maybe just 1cm width does. Along with that, 1cm cross-section touches the ground making 1sq cm. Times that by 2 tires, so 2sq cm. That makes 2sq cm / 0.1 tons = 200sq cm per ton of rubber contact.

That means cyclist has more rubber contact with tarmac than a car does. So doesn't that mean the bike *should* stop faster than a car, but it doesn't, because the brakes do not generate as much stopping force as a car's large brakes do ? Not only that, shouldn't a bike corner faster than a car also? (I saw a test where a sports car can out grips a super(motor)bike).

So what do you think ?

clichty

Well a major factor you omitted is the velocity at which the bike or car is traveling, right? Sure something may have great stopping power but monster difference in stopping time if its going 15 mph vs 75 mph. Also the brake calipers on a car are substantially different than on a bike. Definitely an interesting question but a lot of variables in this equation.

SteelCan

I am not following your math. You get close but your multiplication is faulty. (your suppositions may also be wrong - width of tire contact for both bicycle and car tire)
225x 2cm = 450/car tire
450 x 4 tires = 1800mm 180 SQUARE CM vs. 2sq cm for a bike. That is pretty significant.
More importantly, two wheels rides a LOT differently than 4 - A whole different dynamic of forces. (I am inferring that turns and the changes to inertia are where the contact is most important)

asgelle

I think total contact area equals weight divided by inflation presssure. A 210 lb bike+rider with tires inflated to 105 psi will have 2 sq in contact area. A 3700 lb loaded car with tires inflated to 30 psi will have 123 sq in contact area. Changing inflation pressure changes contact area.

By the way, this is dealt with in detail in "Bicycling Science." There's no need to reinvent this wheel.

JohnnyCyclist

None of my bikes have ABS or traction control, sorry.

AEO

it doesn't matter.
If both are only going in a straight line, then the forces will not involve left/right.

but yeah, I think there is something amiss with the formula.


In particular, I think you are missing weight distribution upon hitting the brakes. We all know the 'weight' shifts to the front under braking. This will deform the tire and change the amount of contact it has with the road and also its μ.

lol headwind

You can slam the brakes in your car pretty hard in a controlled stop, try that on your bike...

DropDeadFred

Who cares

AEO

IMO, it would be easier to do this by comparing either time/distance/acceleration or momentum and kenetic energy.

use the formulas here:
https://www.explainthatstuff.com/motion.html

JohnnyCyclist

I'm more worried about my carbon fork

clichty

Wow lots of cross threading today...

wphamilton

The problem is that braking friction is not directly related to the size of the contact patch.

DropDeadFred

As you should be

mustang1

Yeah you're right. I didn't explicitly enter tire pressure into the equation, but I *kinda* implied it. See on the bike I said it was 1cm width and 1cm length touching the tarmac (due to high pressure) and on car it was 2cm length touching tarmac due to lower air pressure.

I left too many variables out.... my maths is too simple. I was just trying to figure it out.

@AEO: thanks for that link.

mustang1

Funny you mention that. Whenever I slam brakes on the bike, I never hit hard enough on front, and always lock the rear. The rear then always slides slightly to the right (maybe it's the way I sit on the bike?). It's always controlled enough and then I release the rear and apply more brake to the front. Not sure why I do that, I just have this inclination to not hit the front brake too hard in the first instance.

DropDeadFred

90% of my braking comes from the front

ColinL

first thing to fix that problem:

don't touch the rear brake when attempting to stop fast on a road bike. at the limit of braking, your rear wheel will be carrying very close to zero weight, and thus will have very close to zero traction. thus it skids if the brake is applied.

you can skid for fun on MTBs and cruisers, but a road tire will instantly delaminate with even a short skid. not much tread there.

asgelle

Stop. Just stop. If you're interested in how a bicycle deccelerates, read Bicycling Science. You're going off in so many wrong directions, you'll never get there on your own.

clichty

Inclination to not slam the front makes sense, thats like a ritual all 10 year old boys pretty much have to go through. Slam on the front brakes too hard...face into pavement.

StanSeven

I think your formula and suppositions are wrong

nire

Locking the rear wheel is like using the handbrake on a car. If you engage the rear brake too quickly, you will lose rear wheel traction as the braking torque shifts weight to the front wheel. Now you are drifting!

Car tires usually have grooves, so the actual contact area is probably reduced by 40%. But contact area is not really important. The force of friction is proportional to the load (Coulomb Friction). A larger contact patch would just spread the load out. Car tires are wider than bike tires because they need to be stronger to carry the higher load.

A sports car or motorcycle may have a lower center of mass than a bicycle + rider, giving them an advantage in braking because there is a reduced shift of weight to the front.

Phantoj

Agreed... where is he wasting the other 10%?

Phantoj

"contact area is not really important" -- only in physics class! In the real world, of course contact area is important!

...but I agree that bikes have a high C.G. and a short wheelbase -- assuming good pavement, the challenge on a bike is lifting the rear wheel on maximal braking.

wphamilton

If you want to be picky, he left out "with respect to braking force". And that's true, in physics class and in the real world, that braking force is not directly related to the size of the contact area. Counter-intuitive to be sure, but you cannot deny physics.

It comes into play when specific compounds, usually softer for greater braking, will wear more quickly with a smaller contact area. Secondly, generating and dissipating heat, and the amount of inflation pressure are factors which are influenced by selecting a contact area size and particular tire construction.

prathmann

A number of problems with the OP. First is that the contact patch area calculation is way off, esp. for the car tire. Look at a car tire and you'll see that the flattened area at the bottom is much longer than 2 cm - it's many times this long. A rough estimate of the contact patch size can be made based on the tire pressure - a 3000 lb. car being supported by 30 psi tires needs a total contact patch area of about 100 sq. in. (3000 lbs./30 lbs./sq. in. = 100 sq. in.). A cyclist of 200 lbs. (incl. bike) supported by 100 psi tires needs a total contact patch of about 2 sq. in. (In practice there's some variation from this since the tires have some rigidity and therefore the pressure isn't completely uniform across the contact patch area.)

So the cyclist on a road bike actually has a smaller contact patch area per unit weight. But that's not the primary reason for poorer braking compared to a car. Good tires on a good surface give you a maximum deceleration of about 1 G independent of contact patch size. A car driver should be able to get quite close to this figure by modulating his braking to avoid skidding (or letting the ABS do the modulation). To get maximum braking the cyclist must first get his weight down and farther back on the bike than the normal position to avoid lifting the rear wheel and having the bike flip. In normal riding position the cyclist will be limited to about 0.6 G just by the geometry of the center of gravity relative to the front wheel contact position. So he first needs to get his weight farther back and that takes a bit of time before he can use his brake for maximum effectiveness. The cyclist also requires much greater skill to apply the brakes with just the right amount of force for maximum braking without starting to skid. If the motorist brakes too hard and feels a skid starting he can just let up on the brakes for a moment and recover. A cyclist is likely to lose control and start sliding on the ground the moment his front wheel starts to skid - so he's going to be more cautious in applying his front brake (which is the critical one for maximum braking power).

Finally, the smaller contact patch of the cyclist does come into play on non-ideal surfaces. If there are surface imperfections such as loose pebbles, oil drops, leaves, etc. then a small contact patch will result in large changes in the effectiveness of the braking as the tire goes over these. That makes it even harder for the cyclist to maintain just the right level of braking while avoiding a skid with loss of control and balance.