Physics people: help me out with this

Triathlon Forum
1

in another thread there’s a discussion about braking, and jackmott gave the one lucid explanation on that thread of why front brakes are more important than rear brakes (cobb’s was lucid, but was on the subject of aerodynamics, not braking force).

pursuant to that, i’m going to make a statement: your center of mass during hard braking, even on level ground, is somewhere in front of the bike.

my questions are: is my statement above correct, and using the right nomenclature? and how do you quantify this? i think you know what i’m getting at, and this describes why the rear brake is ineffectual: there’s no weight on it during hard braking, so you’ll simply lock up the wheel. but i don’t know if my nomenclature is correct. is it center of mass i’m talking about?

and how would you measure this? center of gravity is a one-dimensional metric (a point on a line) whereas center of mass is a point on an XY axis (it sits some height above the ground). you’d need to know the weight of the bike/rider, and the center of mass at rest, the bike’s wheelbase, and then the velocity of the bike, and the force applied to the front wheel during braking. right? how would one figure out where, in front of the bike, the center of mass sits during braking?

2

I agree with your concept but I am not sure how to put it in correct “physics terms”. Think about it like this: When you are Mtn Biking and going down a steep rugged trail, most riders will drop their seats and actually end up with their butt hanging over behind the seat above and behind the center line of the rear wheel. This in effect is putting your weight as far back as possible. Why? Because it keeps your rear wheel from sliding, exactly like you are proposing for why rear brakes are not as effective as front brakes.

3

Your center of mas (COM) is a geometrical concept that does not change position when you brake unless your body moves or your rear wheel lifts off the ground. What is occurring is that there is a torque being caused by your center of mass being above the ground and the ground being the object that your wheel is acting on to try to slow you down. This torque acts in such a way as to apply more force to the front wheel and take force away from the rear wheel. Mass has not moved just the forces applied to it.

4

yes, this makes sense. nevertheless, if you were to measure the weight on the front v rear wheels during braking, it would be very different than at rest. so is it center of gravity i’m looking for?

5

in another thread there’s a discussion about braking, and jackmott gave the one lucid explanation on that thread of why front brakes are more important than rear brakes (cobb’s was lucid, but was on the subject of aerodynamics, not braking force).

pursuant to that, i’m going to make a statement: your center of mass during hard braking, even on level ground, is somewhere in front of the bike.

my questions are: is my statement above correct, and using the right nomenclature? and how do you quantify this? i think you know what i’m getting at, and this describes why the rear brake is ineffectual: there’s no weight on it during hard braking, so you’ll simply lock up the wheel. but i don’t know if my nomenclature is correct. is it center of mass i’m talking about?

and how would you measure this? center of gravity is a one-dimensional metric (a point on a line) whereas center of mass is a point on an XY axis (it sits some height above the ground). you’d need to know the weight of the bike/rider, and the center of mass at rest, the bike’s wheelbase, and then the velocity of the bike, and the force applied to the front wheel during braking. right? how would one figure out where, in front of the bike, the center of mass sits during braking?
No physics degree here but I think that the center of mass is ABOVE the center line of the two axles , not in front of the bike. This causes the the center of mass to want to lift up and go over the front axle and is counteracted by the weight of the rider & bike so you are unweighting the rear wheel and the front wheel now has more effective weight on it so must now produce more of the stopping force.

JJ

6

Bingo.

AP

7

Dan-
Your falling into the trap of using laymans imprecise terms to try and describe physics concepts. Weight is a force, scales measure forces. The race car term for what you are talking about is % weight transfer. I.E. with a specific setup we experience 80% weight transfer. This would mean that 80% of the total weight of the car is applied to the front wheels during braking with that setup. Due to suspensions the cars actually squat some and the COM of mass does move but due to the fact that we run no suspension and high pressure tires with negligible squish (someone is going to argue with me on the negligible part) there is no movement in the center of mass.

8

Dan-
To add to my original statement and to answer the second part of your question it would be much much easier to calculate the % weight transfer for a given rider on a given bike accelerating in a negative direction (decelerating) then it would be to directly measure it.

9

Dan,
The Motorcycle Safety Foundation has a book called the guide to motorcycling excellence- it provides both physics based calculations and “real world” feel to braking cornering etc. reference chapters 11-14

granted it is geared to motorcycles but the concepts transfer well

k

10

Dan,

This is assuming that in hard braking you are not shifting your weight to the rear of the bike so that the rear wheel/brake can be more involved and engaged in the process. I thought this was rule#1 for hard/emergency braking on a a bike.

11

when in doubt…
F=Ma
center of mass and center of gravity are interchangable. the idea of center of gravity is that its the point at which gravity acts. If you dont change position, it dosent change.
the center of mass isnt moving, like someone previously said. think of a rigid robot on the bike. its body/mass is locked in place. when you apply the brakes, and the mass dosent move, you’re changing the acceleration.
at rest the acceleration is gravity, and thats parallel to a ray coming out of the center of the earth (assuming the earth is a uniformly dense sphere- i know, its not, but thats not what im talkin bout)
when applying the brakes, the mass wants to keep going forward, so you’re “accelerating” it backwards.
Now you have G accelerating you down, and the braking force accelerating you backward- then there is a resultant force… shit. i gotta get back to work
basically your center of gravity is still above the bike
you have momentum
your center of gravity wants to keep going forward, the combined bike/rider weight is well higher than the critical pivot point- which is where your front wheel contacts the ground.
center of gravity keeps moving, pivot point stays the same, and you get flung over the handlebars.
or something like that
but i’m just a structural engineer. the stuff I deal with is mostly stationary. ask the mechanical guys.

12

I’m the mechanical guy, you got it all correct. And stated it clearer than I was doing.

13

Wikipedia to the rescue again.

Go to “Bicycle and motorcyle dynamics.” Lots of fascinating stuff there, but the appropriate section copied from Wikipedia here:

Longitudinal stability http://upload.wikimedia.org/wikipedia/en/thumb/d/d9/Stoppie.jpg/180px-Stoppie.jpg http://upload.wikimedia.org/skins/common/images/magnify-clip.png A motorcycle performing a stoppie.
Mechanical analysis of the forces generated by a bike with a wheelbase L and a center of mass at height h and halfway between the wheels, with both wheels locked, reveals that the normal (vertical) forces at the wheels are: http://upload.wikimedia.org/math/a/6/8/a68232eb3fa15d2f1a7657001522e8e2.png for the rear wheel and http://upload.wikimedia.org/math/7/2/e/72e01cf412c8d0fb8f2e2bd94a0454c6.png for the front wheel,
while the frictional (horizontal) forces are simply F**r = μN**r for the rear wheel and F**f = μN**f for the front wheel, where μ is the coefficient of friction, m is the mass, and g is the acceleration of gravity. Therefore, if http://upload.wikimedia.org/math/4/b/e/4befbdb320851757a1776bddea92fc63.png
then the normal force of the rear wheel will be zero (at which point the equation no longer applies) and the bike will begin to flip forward over the front wheel.
The coefficient of friction of rubber on dry asphalt is between 0.5 and 0.8. Using the lower value of 0.5, and assuming the center of mass height is greater than or equal to the wheelbase, the front wheel can generate enough stopping force to flip the bike and rider forward over the front wheel.
On the other hand, if the center of mass height is less than half the wheelbase and at least halfway towards the rear wheel, as is true, for example on a tandem or a long-wheel-base recumbent, then, even if the coefficient of friction is 1.0, it is impossible for the front wheel to generate enough braking force to flip the bike. It will skid unless it hits some fixed obstacle, such as a curb.
In the case of a front suspension, especially telescoping fork tubes, this increase in downward force on the front end may cause the suspension to compress and the front end to lower. This is known as brake diving. A riding technique that takes advantage of how braking increases the downward force on the front wheel is known as trail braking.
edit] Front wheel braking
The limiting factors on the maximum deceleration in front wheel braking are: the maximum, limiting value of static friction between the tire and the ground,the kinetic friction between the brake pads and the rim or disk,pitching (of bike and rider) over the front wheel.
For an upright bicycle on dry asphalt with excellent brakes, pitching will probably be the limiting factor. The combined center of mass of a typical upright bicycle and rider will be about 60 cm (23.6 in) back from the front wheel contact patch and 120 cm (47.2 in) above, allowing a maximum deceleration of 0.5 g (4.9 m/s² or 16 ft/s²). If the rider modulates the brakes properly, however, pitching can be avoided. If the rider moves his weight back and down, even larger decelerations are possible.
Front brakes on many inexpensive bikes are not strong enough so, on the road, they are the limiting factor. Cheap cantilever brakes, especially with “power modulators”, and Raleigh-style side-pull brakes severely restrict the stopping force. In wet conditions they are even less effective.
Front wheel slides are more common off-road. Mud, water, and loose stones reduce the friction between the tire and trail, although knobby tires can mitigate this effect by grabbing the surface irregularities. Front wheel slides are also common on corners, whether on road or off. Centripetal acceleration adds to the forces on the tire-ground contact, and when the friction force is exceeded the wheel slides.
Of course, the angle of the terrain can influence all of the calculations above. All else remaining equal, the risk of pitching over the front end is reduced when riding up hill and increased when riding down hill.

GNU license to copy this text: http://en.wikipedia.org/wiki/Wikipedia:Text_of_the_GNU_Free_Documentation_License

14

I admit. I half assed it.
If you really want to know…
The bike is like a simply supported beam, supported at both wheel contact points, typically around 3’ apart
The forces acting on it are as I described before, gravity, and the acceleration due to braking.
Moments are the key (moment=> momentum) which is a force times a moment arm.
F=Ma. Mass being the rider’s mass in kg. A being whatever gravity is, and whatever the braking force is.
Your moment arms will be the distance from the front wheel’s contact, back to the rider CG, say… 1.5’ , and the distance from the ground to the rider CG, say 4’.
now talk about newton- every action has an equal and opposite reaction.
The forces acting on your CG are down and backward- so the forces acting on the simply supported beam are forward and up.
at rest, you could say that the two up forces are equal, if your CG is equidistant between the front and rear wheel.
a 100lb (or slugs if you wanna get technical) rider is going to have 50lb at the front wheel and 50 at the rear wheel. sum of the forces in the Y direction equals zero. 50+50-100=0
there are no forces in the X direction. you’re at rest.

Braking is a diff story.
the braking force is pushing you back. so NOW you have X forces. you still weigh 50lb, the total between the reaction of your front and rear wheel in the Y direction is still 100lb. but now, the front and rear wheel no longer have it equally distributed.
IF the braking acceleration and gravity are the same…
the sum of the moments about each point must equal zero to be “static”. Moments in this case will be in lbft.
the moment about the rear wheel contact is the sum of…
-rider mass times distance in Y direction from rear wheel contact = 400lbft clockwise (about the rear wheel contact)
-rider mass times distance in X direction from rear wheel contact = 150lbft clockwise (about the rear wheel contact)
-reaction at front wheel times distance in X direction from rear wheel contact = ? counterclockwise (about the rear wheel contact)(because the force is acting upwards)
550lbft clockwise + 550lbft counterclockwise = 0
so the moment from the reaction at the front wheel is 550lbft
its 3’ away from the rear wheel, so the force acting upwards is 550/3=183lb.
whats that mean? without either changing the center of gravity, or somehow resisting the other 83lbs with the rear wheel, that rider is goin over the handlebars.
sum of the forces in the Y direction should still be zero, so 183lb up front means you need 83 pulling down on the rear wheel to avoid flipping.
maybe my #s werent realistic. but the math is there. its all about moments and forces equalling zero.
just hang your ass off the back of the seat if you get into a hairy braking situation. front brake is hugely important. dont forget that.
any questions? feel free to pick this apart. i’m going home soon, and going out, and then going out tomorrow. i wont be around to defend myself. have at it.
slowman- feel free to PM me if you really care to go deeper, or find real numbers, or figure wtf im talking about

15

just hang your ass off the back of the seat if you get into a hairy braking situation. front brake is hugely important. dont forget that.

You seem to know far more about the physics side of this than me, but above, is the bottom line and what I said back in post#10 - key thing to keep the back-end of the bike engaged in the braking process and to keep the forces balanced is shift weight to the rear of the bike.

16

From a geeky site: http://www.nationmaster.com/encyclopedia/Bicycle-and-motorcycle-dynamics#Braking says that it’s a relationship between the distance between the wheel axles and the height of the center of mass, along with a coefficient of friction. It can be impossible for a rider to go over the bars:

On the other hand, if the center of mass height is less than half the wheelbase and at least halfway towards the rear wheel, for example a tandem or a long-wheel-base recumbent, then even if the coefficient of friction is 1.0, it is impossible for the front wheel to generate enough braking force to flip the bike. It will skid unless it hits some fixed obstacle such as a curb.

Note that neither velocity nor mass figure into the final calculation: it’s all about the geometry and the coefficient of friction, and the combination that results in the rear wheel lifting off the ground. You can lift the rear wheel off the ground at very low speed if your COM is high enough (try (carefully) standing on the bike and applying the front brake at low speed).

17

Are you talking about Center of Resistance?

18

in another thread there’s a discussion about braking, and jackmott gave the one lucid explanation on that thread of why front brakes are more important than rear brakes (cobb’s was lucid, but was on the subject of aerodynamics, not braking force).

pursuant to that, i’m going to make a statement: your center of mass during hard braking, even on level ground, is somewhere in front of the bike.

my questions are: is my statement above correct, and using the right nomenclature? and how do you quantify this? i think you know what i’m getting at, and this describes why the rear brake is ineffectual: there’s no weight on it during hard braking, so you’ll simply lock up the wheel. but i don’t know if my nomenclature is correct. is it center of mass i’m talking about?

and how would you measure this? center of gravity is a one-dimensional metric (a point on a line) whereas center of mass is a point on an XY axis (it sits some height above the ground). you’d need to know the weight of the bike/rider, and the center of mass at rest, the bike’s wheelbase, and then the velocity of the bike, and the force applied to the front wheel during braking. right? how would one figure out where, in front of the bike, the center of mass sits during braking?
One can estimate this pretty well if you separate the various components. Each component has a pretty easily estimated mass and the center of mass of that component is pretty easily estimated. Head, upper arms, forearms, chest, abdomen, legs. The center of mass would be the weighted average of all these masses on the x, y and z axis. So, the center of mass will depend upon the position of the rider. I was taught to shift my CM back during emergency stops, not so much to help the back wheel brake harder but to keep from going over the handlebars, which is much easier to do the closer the CM is to being over the front wheel.

19

The difference here (if I am understanding your problem correctly) is that weight is a force, not a mass. The weight on the front will change during braking because you have a torque applied to it, however, your center of mass will not move. Finding the center of mass is not that difficult to do.

Example

Bike COM: x=2, w=10lbs
Body COM: x=4, w=200lbs

Xcom = ((210)+(4200))/(200+10)

do the same thing for Y and Z

20

Dan,

Wikipedia has a simple and excellent discussion of center of mass. Here is a link

As several people pointed out, center of mass is a static concept.

In that article there are several links to the dynamics of car (and other vehicle handling issues).

Included in that article is a discussion of weight distribution

And more interestingly … car handling which an extensive discussion of breaking and weight distribution.

Alan