The 40/60 split - - - setback and stuff.

[false_Aest]

Hey,

In the setback seatpost thread it was suggested that one could figure out his setback by throwin the bike in a trainer and a scale under the front wheel.

Looking for something around a 40/60 weight distribution.

I tried to rock the 40/60 split on the trainer last night out of curiosity. Maybe my femurs are meant for the Lollypop Guild but I found that I'd need to throw my saddle back 2-2.5cm and then swap my stem from a 120 to a 100. (When I calculated my current weight split, I'm coming up with 44/56).

Eric, in no way do I intend this to be a "calling out." Instead, I'm curious as to what others think about this idea . . . Mr. Kirk? TiDesigns? would you care to offer your input.

Also, if there happens to be a "trend" for a 40/60 split is that for a specific uhm . . . demographic (casual vs pro, 1, 2)?

Thanks!

[Richard]

I've always aimed for a 55/45 ratio. It is what I was told was racer ideal low those many years ago.

I also would be a bit cautious about using the trainer method to figure out the ratio. We used to use two scales, one under each wheel and another person to hold the bike and rider upright.

[Dave]

To be totally accurate, I'd block the rear wheel up so the bike is sitting level when the scale is under the front wheel. Place the bike near a wall so you can use an elbow to lean against if needed. Weigh the bike and rider first, then figure out the 42-45% range and look for that, while one the bike. Use the agressive cornering position when measuring the weigh balance. There's no need for two scales.

IMO, 40% on the front is on the light side.

[ergott]

60/40 isn't set in stone. I've heard different numbers for different builders in the range or 60/40 to 55/45.

[AndrewS]

I have to say that this whole methodology seems a little far out.

Position is primarily about good ergonomics, I had thought. But this weight distribution thing implies that a guys who goes from having a fat butt and skinny arms is going to need to change his bike position after spending a lot of time at the gym. His butt shrinks, his forearms bulge, so now his stem is too long?????

Aside from the relative difficulty of getting the measurements right and that moving your 10 lbs. head up and down will change them, I can't see fighting one's best, most comfortable position to try and satisfy a largely theoretical weight distribution. Doesn't bicycle frame design already takes most of this stuff into account?

How many people with a decent fit on a racing bicycle fail to have an adequate weight distribution?

[false_Aest]

I have to say that this whole methodology seems a little far out.

Position is primarily about good ergonomics, I had thought. But this weight distribution thing implies that a guys who goes from having a fat butt and skinny arms is going to need to change his bike position after spending a lot of time at the gym. His butt shrinks, his forearms bulge, so now his stem is too long?????

Aside from the relative difficulty of getting the measurements right and that moving your 10 lbs. head up and down will change them, I can't see fighting one's best, most comfortable position to try and satisfy a largely theoretical weight distribution. Doesn't bicycle frame design already takes most of this stuff into account?

How many people with a decent fit on a racing bicycle fail to have an adequate weight distribution?


Yeah I started thinking along these lines during my ride this morning.

Thanks for everyone's eventual input.

[AndrewS]

With a custom bike this makes more sense. A builder can choose what top tube/front center and stem length combination puts the front wheel at the best balance point. That's what Eric refers to in his last post.

But as a fitting scheme to an existing bike - yikes.

[ergott]

With a custom bike this makes more sense. A builder can choose what top tube/front center and stem length combination puts the front wheel at the best balance point. That's what Eric refers to in his last post.

But as a fitting scheme to an existing bike - yikes.

It can also work if you are trying to see what stock frames fit you before you purchase. Try for a 25mm offset or the stock seatpost (if it comes with one) and try to get your reach (saddle to bars) with an acceptable length stem/stack. The hands usually end up directly above the front hub (note, I'm not saying you should look at the bars and see where the hub ends up).

If you can't come close to this, try a different frame geo.

I do emphasize that this is approximate and a guideline.

Don't make yourself fit the bike, find the bike that fits you.

[MadRocketSci]

rough calcs...maybe this is common knowledge, was just thinking a little bit and thought i'd throw this out there...

assuming front center to rear center ratio of ~60/40, a 55/45 weight distribution puts your cg 3-5 cm in front of the bb. so if you're cornering, you will be in a balanced position with all your weight supported by your outside foot (no weight on the seat or bars) with your leg parallel to the seat tube, since 17(=crank length)*sin(17(=seat tube angle wrt vertical)) is about 5.

Braking will change this distribution. Why 55/45? Why not 50/50? maybe to ensure more weight on the rear to help keep it from sliding out....or maybe to account for braking during cornering...

[false_Aest]

rough calcs...maybe this is common knowledge, was just thinking a little bit and thought i'd throw this out there...

assuming front center to rear center ratio of ~60/40, a 55/45 weight distribution puts your cg 3-5 cm in front of the bb. so if you're cornering, you will be in a balanced position with all your weight supported by your outside foot (no weight on the seat or bars) with your leg parallel to the seat tube, since 17(=crank length)*sin(17(=seat tube angle wrt vertical)) is about 5.

Braking will change this distribution. Why 55/45? Why not 50/50? maybe to ensure more weight on the rear to help keep it from sliding out....or maybe to account for braking during cornering...

that's some mad rocket science that I didn't get.

[Dave]

If you place all of your weight on the downside pedal, it will place even more weight on the front. I do NOT measure the weight balance in that position. I often ride mountains descents with a series of right and left turns, closely spaced. I DO (edited) change from right to left side pedal down, so I don't touch the pavment, but I never bother weighting the pedal because I don't need to with 45% of the weight on the front.

[MadRocketSci]

yes, bad explanation made more difficult without pictures.

I think I was trying to say that to achieve a 55/45 weight balance, your cg needs to be in front of the bb, which it would if you were successfully cornering with all your weight on your downside foot extended such that your cranks were along the seat tube axis, and no weight on your butt or hands. If all you have to do is just unweight your butt without sliding backwards or forwards on your saddle, then your bike is ideally set up for cornering (assuming 55/45 is where it's at).

BTW, I do bother with switching legs on those linked turns I used to do down Kings and Page Mill near Woodside. Maybe it's a habit kept from skiing.

If you moved your butt back about 3-5 cm you would be more 60/40 dist. Does it make a difference? perhaps if you're doing some heavier braking and loading your front wheel more, it might help.

[rice rocket]

If you moved your butt back about 3-5 cm you would be more 60/40 dist. Does it make a difference? perhaps if you're doing some heavier braking and loading your front wheel more, it might help.
What's your endgame here? Are you trying to achieve higher cornering stability/grip by hitting some weight distribution number in turns?

How about doing what's comfortable, so you don't wreck trying to achieve yoga poses while cornering at full tilt? :p

[MadRocketSci]

assume 55/45 back/front weight dist is ideal...this is what bike designers shoot for apparently

you get this dist while cornering if you are bearing most of your weight on your downhill leg, extended to the max, ie, with the crank arms aligned with seat tube axis. This is how many corner, no yoga poses here.

hopefully this happens in a comfortable natural position...if so then its a good bike for u as far as weight distribution goes.

if it doesn't then maybe bike is not for you.

that's all....

[Dave]

I had to edit my post about the right/left pedal being down. Of course, I change from right to left side pedal being down when the curves alternate direction. I just don't need to rely on weighting the pedal to get enough weight on the front tire. It's there, just due to my low torso position.

[AndrewS]

Just a thought: A 45/55 split on level ground is a 50/50 split on a steep downhill grade. I wonder how that factors into things?

[flickwet]

Back in the day when Schwinns were sold at Schwinn stores, simplex was a lousy deralleur and all Italian Bikes were made in Italy, I remember something about: butt on the seat, hands in the drops; if the stem is right the flat of the bar will line up over the front axle when looking down, but back then a size 57 was a seat tube measurements, now its effective tt, and top tubes were always "under square in measurement to st, I have always still used that "look down" as a quick way to getting a base line on stem length then adjusting seat set back according to what works for the rider.

[MadRocketSci]

I had to edit my post about the right/left pedal being down. Of course, I change from right to left side pedal being down when the curves alternate direction. I just don't need to rely on weighting the pedal to get enough weight on the front tire. It's there, just due to my low torso position.

Three places that can bear weight on the bike, hands, feet, butt. I would contend that, assuming little weight on the hands, that if you are having the seat support some of your weight then you have a higher rear/front weight dist ratio.

This can be seen by drawing a free body diagram, using sum of torques = 0, upward force at bike seat must be balanced by upward force at pedal, which necessitates the cg being behind the cg value where 55/45 weight dist is achieved.

[MadRocketSci]

Just a thought: A 45/55 split on level ground is a 50/50 split on a steep downhill grade. I wonder how that factors into things?

the angle of the road will not change weight distribution....the component of gravity along the road just causes the rider to accelerate. The components of force perpendicular to the road remain in the same distribution.

Again, the free body diagram will help visualize this...

http://en.wikipedia.org/wiki/Free_body_diagram

[David Kirk]

In reading this thread I'm not sure I understand what it's about at this point, Nonetheless I'll throw my two cents in.

The rider's fit is the ride's fit and it's a matter of putting the 3 contact points in the right places relative to each other in space. The fit is rarely, if ever, changed on a given bike to influence fore/aft weight distribution. Fit comes first.

If the rider has a stock bike that he has been fit to then there is no effective adjustment of the weight distribution to be had without changing the fit. If the bike is being designed and built for that rider then the weight distribution can be taken into account and optimized for that rider, on that bike.

Or something like that.

dave

[sevencyclist]

In reading this thread I'm not sure I understand what it's about at this point, Nonetheless I'll throw my two cents in.

The rider's fit is the ride's fit and it's a matter of putting the 3 contact points in the right places relative to each other in space. The fit is rarely, if ever, changed on a given bike to influence fore/aft weight distribution. Fit comes first.

If the rider has a stock bike that he has been fit to then there is no effective adjustment of the weight distribution to be had without changing the fit. If the bike is being designed and built for that rider then the weight distribution can be taken into account and optimized for that rider, on that bike.

Or something like that.

dave
I suppose with frame design, the three contact points can be floated around. But on an existing frame, the pedals/crankset is locked in, so the other two contact points will need to follow.

With that said, if saddle is moved forwards as in time trial position, then there will be slight angle changes that influence weight distribution, and the bar/stem might need to be adjusted to compensate for the position difference. Thus, the three points will no longer be constant to each other.

Similarly, when the bike is on an incline, the fore/aft on saddle might change again and the weight distribution changes, which might alter the power output as well as handling. Must be hard to design a bike to anticipate those changes.

[AndrewS]

the angle of the road will not change weight distribution....the component of gravity along the road just causes the rider to accelerate. The components of force perpendicular to the road remain in the same distribution.

Again, the free body diagram will help visualize this...

http://en.wikipedia.org/wiki/Free_body_diagram
I don't think that applies the way you're thinking. If I lift the back wheel, the front wheel on the scale will show more weight.

In a frictionless system, the acceleration may offset the angle intially, but on normal grades the bike quickly reaches terminal velocity and the vectors are the same as if it were stopped on the downhill. Same reason skydivers only experience a few seconds of "weightlessness" before gravity asserts itself again on their inner ears - they stop accelerating.

Similarly, when the bike is on an incline, the fore/aft on saddle might change again and the weight distribution changes, which might alter the power output as well as handling. Must be hard to design a bike to anticipate those changes.Only if the current design isn't doing the job, which it pretty much is.

Keith Bontrager came up with a CG based fit system:
http://www.sheldonbrown.com/kops.html
The funny part is that he ended up with the same seat angles, 72.5 - 74, that are already in use. I think geometry can be used to refine specific characteristics of ride, but standard road bike geometry's evolution has taken care of many of the concerns of both fit and handling without requiring builders to go back to square one when a question of physics comes up.

[mister]

guys. dave is saying fit comes first.
set your bike up so you are in the position you need.
the weight distribution falls where it falls. the bike will ride well or it won't. don't screw up your position to try to change the way a bike rides.

[54ny77]

60% of me wants to eat like a pig and drink beer, 40% feels guilty about that and prefers exercise.

At the moment, majority rules!

That's my fit method and I'm sticking to it. :D

[David Kirk]

guys. dave is saying fit comes first.
set your bike up so you are in the position you need.
the weight distribution falls where it falls. the bike will ride well or it won't. don't screw up your position to try to change the way a bike rides.

I wish I'd have said it so succinctly.

Dave

[false_Aest]

Yeah, this is what I figured but thought it was worth checking out my weight distribution with my current set-up. The result was just a tiny bit of confusion

Thanks for the input an clarification.

[ergott]

Don't make yourself fit the bike, find the bike/builder that fits you.

What this guy said.

[sevencyclist]

Don't make yourself fit the bike, find the bike that fits you.
Better yet, find a builder who understands the handling dynamics and build a bike to fits you.

[ergott]

Better yet, find a builder who understands the handling dynamics and build a bike to fits you.

Fixed it.

[MadRocketSci]

I don't think that applies the way you're thinking. If I lift the back wheel, the front wheel on the scale will show more weight.

It shouldn't. Not according to Newton's laws:

sum of forces = mass*accel
sum or external torques on bike = 0

angle falls out of the torque equations (except in the vertical case) and does not affect the ratio of rear to front wheel force.


In a frictionless system, the acceleration may offset the angle intially, but on normal grades the bike quickly reaches terminal velocity and the vectors are the same as if it were stopped on the downhill. Same reason skydivers only experience a few seconds of "weightlessness" before gravity asserts itself again on their inner ears - they stop accelerating.

terminal velocity due to aerodynamics is pretty high. For humans in freefall it is around 120 mph. I've never tested it on a bike, not sure why it is relevant to weight distribution.

[MadRocketSci]

guys. dave is saying fit comes first.
set your bike up so you are in the position you need.
the weight distribution falls where it falls. the bike will ride well or it won't. don't screw up your position to try to change the way a bike rides.

You want both in a bike. If the bike fits but handles like crap you need to find a new bike unless you only ride in straight lines and gentle turns. You want the bike that "fits" and simultaneously puts your cg at the right place between the wheels.

[alembical]

now I am confused....

Does a jet on a treadmill ever take off?
What happens to a hummingbird hovering inside your car when you accelerate?

:fight:

[false_Aest]

now I am confused....

Does a jet on a treadmill ever take off?
What happens to a hummingbird hovering inside your car when you accelerate?

:fight:


The jet will take off because its wheels are not its means of propulsion, its engines are.

The hummingbird will move toward the back of the car. Similarly if you stop suddenly the hummingbird will crash into the window. So will cigarette smoke. A balloon will move toward the back of the car because of blah blah blah air something or other.

[AndrewS]

It shouldn't. Not according to Newton's laws:

sum of forces = mass*accel
sum or external torques on bike = 0

angle falls out of the torque equations (except in the vertical case) and does not affect the ratio of rear to front wheel force.



terminal velocity due to aerodynamics is pretty high. For humans in freefall it is around 120 mph. I've never tested it on a bike, not sure why it is relevant to weight distribution.
In the first case, if you were right, there would be no such thing as a center of gravity. But when you put two legs of anything on two scales, then tip it toward one scale or the other, the weight shifts. Eventually, the weight on the high scale equals zero and the low scale bears the entire load. Look in you science book at the section about levers. A non-accelerating bicycle is exactly the same.

You reach terminal velocity every time you coast down a hill and stop accelerating. That's when the sum of drag on the bicycle equals the force pulling the bike down against the plane. Steeper hills and better aerodynamics raise terminal velocity.

When you reach that equilibrium and stop accelerating the vectors of gravity act just like you do when you're stopped, so it pertains to a discussion of a bicycle on a downhill. When accelerating those vectors are shifted, which was the point I thought you were getting at when you mentioned acceleration.

[David Kirk]

The jet will take off because its wheels are not its means of propulsion, its engines are.

The hummingbird will move toward the back of the car. Similarly if you stop suddenly the hummingbird will crash into the window. So will cigarette smoke. A balloon will move toward the back of the car because of blah blah blah air something or other.



OK.......... here's a good one. If you are driving in a car with the windows up and you have a helium balloon on a string and you make a hard right turn what does the balloon do?

a) it swings to the left just like you think it would.

b) it swings to the right just to confuse you.

c) it bobs up and down wildly and then assplodes.

d) it does nothing.



What say you physics experts?

Dave

[Craig Ryan]

B
The heavier air moves to the outside of the turn. Same thing when you accelerate or brake.

[54ny77]

If a train left Chicago traveling at....

[Mark McM]

I don't think that applies the way you're thinking. If I lift the back wheel, the front wheel on the scale will show more weight.

Not quite. If the rear wheel is higher than the front wheel (as on the sloped surface of a hill), then you must apply braking to the wheels to keep the bike from rolling downhill. This adds an extra force to the system, which acts to shift the weight distribution.

In a frictionless system, the acceleration may offset the angle intially, but on normal grades the bike quickly reaches terminal velocity and the vectors are the same as if it were stopped on the downhill. Same reason skydivers only experience a few seconds of "weightlessness" before gravity asserts itself again on their inner ears - they stop accelerating.

But the difference is that when braking, the deceleration force is applied at the wheel/ground contact point, which is far below the center of gravity. This induces a torque on the system, which must be countered by shifting weight from the rear wheel to the front wheel.

When coasting at terminal velocity, the center of the deceleration force (air resistance) is much closer to the center of gravity, and therefore there is little torque created by the deceleration force, and so little change in weight distribution between the wheels.

Keith Bontrager came up with a CG based fit system:
http://www.sheldonbrown.com/kops.html
The funny part is that he ended up with the same seat angles, 72.5 - 74, that are already in use. I think geometry can be used to refine specific characteristics of ride, but standard road bike geometry's evolution has taken care of many of the concerns of both fit and handling without requiring builders to go back to square one when a question of physics comes up.

The Keith Bontrager article on fitting based on CG only considers the position of the rider's CG with respect to the pedals/cranks. It does not consider weight distribution between the wheels.

[AndrewS]

Hi Mark,

I think you are confusing active braking (deceleration) with locked brakes on something static (parking brakes on a car, the feet of a ladder). When you decelerate, it of course shifts weight toward the front wheel, downhill or on level ground. That's why the front brake is more effective.

If you take something with a steady velocity (stopped on not accelerating - motion is relative) and raise one of its legs (or wheels), the weight shifts toward the lower. That's true whether it is a ladder set on an uphill, a parked car, or a bicycle that has topped out (terminal) at 38 mph on a downhill. This is why locking up the front brake on a downhill is more dangerous than on a flat - your CG is already tipped toward the front wheel, so it takes less deceleration to fling you over the bars.

You and Madrocket are over thinking this stuff (or underthinking). You know this stuff in the real world, or you'd be falling off ladders and your bikes all the time. Don't want to believe me OR your inner ear? Hang something from your rearview mirror, then drive down and park on a steep hill. You'll find the angle of the dangle is identical whether you're going a steady 35 mph or parked.

I only brought up the downhill example to point out that having exact weight distribution numbers didn't really make sense - slope, body position, head angle, etc can throw them wildly off.


I realize the Bontrager article was about a different CG issue. I put it up just to make the point that modern frame geometry really doesn't need new types of analysis to work right - all of the "issues" have been accounted for with a century of trial and error.

[MadRocketSci]

In the first case, if you were right, there would be no such thing as a center of gravity. But when you put two legs of anything on two scales, then tip it toward one scale or the other, the weight shifts. Eventually, the weight on the high scale equals zero and the low scale bears the entire load. Look in you science book at the section about levers. A non-accelerating bicycle is exactly the same.

You reach terminal velocity every time you coast down a hill and stop accelerating. That's when the sum of drag on the bicycle equals the force pulling the bike down against the plane. Steeper hills and better aerodynamics raise terminal velocity.

When you reach that equilibrium and stop accelerating the vectors of gravity act just like you do when you're stopped, so it pertains to a discussion of a bicycle on a downhill. When accelerating those vectors are shifted, which was the point I thought you were getting at when you mentioned acceleration.

AndrewS,

I believe you are the one who is over/underthinking this whole thing. No offense, my head hurts trying to follow the logical paths you are laying out.

You might be talking about the dynamics of raising one end of something wrt the other end. In that case, yes, the fixed end could show higher forces due to the fact that it is constraining a dynamic system, ie, one that is moving, such that the act of lifting the back end causes the cg of the system to move higher and also imparts and angular velocity on the system. (BTW, you can perform the lift of a bike's back wheel without having to apply braking to the front if you carefully ensure the point you are lifting follows a circular arc wrt the front tire contact point).

Here is what I am talking about and it is a very simple application of Newton equations. If you feel the need challenge my analysis, then do it from the fundamental laws, not a "since this happens, then it applies to this" type argument...since the first part of the statement is also in question.

Here goes:
Bike + rider on level ground, moving at constant velocity.
acceleration along and perpendicular to ground = 0

therefore, sum of forces in both axes (parallel to ground, perp to ground) = 0

forces on bike + rider system:
gravity acting on system CG, call it W, down
upward force on rear tire, call it Fr, up
upward force on front tire, call it Ff, up
ignore aerodynamics - it is not relevant here

Newton's law in translational motion:
sum of forces = 0
define "up" as positive

Fr + Ff - W = 0
therefore Fr + Ff = W

We agree? hope so.

Newton's law in rotation:
sum of Torques = change in angular momentum = 0
the system is not rotating so this is true
define x as the horizontal distance from system CG to rear wheel contact point with ground
ditto y as distance from system CG to forward wheel contact point
define positive as from rear to front wheel

sum of torques about CG = Ff*y - Fr*x = 0
therefore Ff*y = Fr*x
Fr/Ff = y/x = weight distribution rear/front ratio

assume this is 55/45

then y/x = 55/45

for my 56 square legend ti
front center = 59.5
chainstay = 41.5

ratio approx 60/40 - therefore, since y < 60 the system cg is in front of bb by about 5 cm. Note the system CG is not the same as rider cg, as the bike cg is part of it.

We all together still?

Question: how does Fr/Ff change given a bike going down a constant slope?

Answer: we still apply newton's laws to axes parallel and perpendicular to road

Gravity still acts straight down, and therefore has components parallel and perpendicular to the road

Define "theta" as angle between horizontal and road...ie, 10% for Alpe d'Huez

total force of gravity = Fg = mass*g, (g = constant = 9.8 m/s2 or 31.74 ft/s2)

Force of gravity parallel to road = Fg*sin(theta)
Force of gravity perpendicular to road = -Fg*cos(theta)

Force of ground on wheels
= 0 parallel to road
= Fr and Ff in direction perpendicular to road, though values will be different from level ground case...we are interested in ratio

Ready to apply newton's laws

Translational motion, parallel to road
sum of forces in this direction = accel in this direction

Fg*sin(theta) - aerodynamic drag = (mass of system)*(accel of system)

This just tells you how fast you will accelerate down the hill...
(aside: aero drag = Cd*frontal area*0.5*density*velocity^2...but who cares here)

No insight gained on weight dist from this axis

Translational motion, perpendicular to road
since we are not going through the ground, the accel in this direction is zero

Fr+Ff-Fg*cos(theta) + aerodynamic lift = 0
forget aerodynamic lift, it is small and you'd need to know the center of pressure which complicates your analysis but not the final result...we're not talking missiles flying at mach 5 here...

therefore,
Fr+Ff = Fg*cos(theta)

need one more equation to solve for Fr, Ff - look at rotational motion

Rotational motion
Gravity produces no torque on system since it acts through CG
x, y are defined as before
Torque due to ground force on rear wheel = -Fr*x (since Fr is perp to x)
Torque due to ground force on front wheel = Ff*y ( "" "" "")

sum of torques = 0 (since we are not rotating at all)
Ff*y - Fr*x = 0
Ff*y = Fr*x
Fr/Ff = y/x as before <- result

QED

We get this result because gravity acts through the CG and therefore does not add any torque to the system. On the other hand, braking does add a torque on the system and hence changes the Fr/Ff ratio.

Now if there is still a debate to this, please reference the above and tell me how I have screwed up the analysis based on fundamental principles (hey, it's possible)...the worlds of math, physics and engineering are full of counterintuitive results that require careful application of fundamental principles to accurately predict and explain. Hand waving arguments (a technical term) are not a basis of proof....

[e-RICHIE]

did you get this from my frame order form atmo?





Here goes:
Bike + rider on level ground, moving at constant velocity.
acceleration along and perpendicular to ground = 0

therefore, sum of forces in both axes (parallel to ground, perp to ground) = 0

forces on bike + rider system:
gravity acting on system CG, call it W, down
upward force on rear tire, call it Fr, up
upward force on front tire, call it Ff, up
ignore aerodynamics - it is not relevant here

Newton's law in translational motion:
sum of forces = 0
define "up" as positive

Fr + Ff - W = 0
therefore Fr + Ff = W

We agree? hope so.

Newton's law in rotation:
sum of Torques = change in angular momentum = 0
the system is not rotating so this is true
define x as the horizontal distance from system CG to rear wheel contact point with ground
ditto y as distance from system CG to forward wheel contact point
define positive as from rear to front wheel

sum of torques about CG = Ff*y - Fr*x = 0
therefore Ff*y = Fr*x
Fr/Ff = y/x = weight distribution rear/front ratio

assume this is 55/45

then y/x = 55/45

for my 56 square legend ti
front center = 59.5
chainstay = 41.5

ratio approx 60/40 - therefore, since y < 60 the system cg is in front of bb by about 5 cm. Note the system CG is not the same as rider cg, as the bike cg is part of it.

We all together still?

Question: how does Fr/Ff change given a bike going down a constant slope?

Answer: we still apply newton's laws to axes parallel and perpendicular to road

Gravity still acts straight down, and therefore has components parallel and perpendicular to the road

Define "theta" as angle between horizontal and road...ie, 10% for Alpe d'Huez

total force of gravity = Fg = mass*g, (g = constant = 9.8 m/s2 or 31.74 ft/s2)

Force of gravity parallel to road = Fg*sin(theta)
Force of gravity perpendicular to road = -Fg*cos(theta)

Force of ground on wheels
= 0 parallel to road
= Fr and Ff in direction perpendicular to road, though values will be different from level ground case...we are interested in ratio

Ready to apply newton's laws

Translational motion, parallel to road
sum of forces in this direction = accel in this direction

Fg*sin(theta) - aerodynamic drag = (mass of system)*(accel of system)

This just tells you how fast you will accelerate down the hill...
(aside: aero drag = Cd*frontal area*0.5*density*velocity^2...but who cares here)

No insight gained on weight dist from this axis

Translational motion, perpendicular to road
since we are not going through the ground, the accel in this direction is zero

Fr+Ff-Fg*cos(theta) + aerodynamic lift = 0
forget aerodynamic lift, it is small and you'd need to know the center of pressure which complicates your analysis but not the final result...we're not talking missiles flying at mach 5 here...

therefore,
Fr+Ff = Fg*cos(theta)

need one more equation to solve for Fr, Ff - look at rotational motion

Rotational motion
Gravity produces no torque on system since it acts through CG
x, y are defined as before
Torque due to ground force on rear wheel = -Fr*x (since Fr is perp to x)
Torque due to ground force on front wheel = Ff*y ( "" "" "")

sum of torques = 0 (since we are not rotating at all)
Ff*y - Fr*x = 0
Ff*y = Fr*x
Fr/Ff = y/x as before <- result

QED

[MadRocketSci]

Yes...royalty checks to you and Isaac's estate are in the mail...

[AndrewS]

Madrocket,

You keep referring to an accelerating body, not one at fixed velocity on a downslope. You include drag, but then only as a limit on the rate of downhill acceleration, not as a stop to acceleration and velocity limit. When that point is reached, the object can't tell whether it is rolling down a hill, rolling on a sloped treadmill, or stopped. That's when the bicycle becomes no different than a ladder.

Do you now understand that terminal velocities of objects running down slopes are lower than freefalling objects, for the same reason that their acceleration is lower?

The only argument I can understand against my point (which I don't believe you were making, but could be wrong) is that aerodynamic drag at terminal velocity is countering CG, thereby pushing the rear wheel back against the ground and taking the weight off the front wheel. Similar to a ladder on a too steep slope being kept from falling by a stiff breeze. This probably works, but not precisely because the aerodynamic center of rider and bike might be quite different than CG. A higher aero center than CG would increase the rear wheel weighting compared to static, and vice versa.

Of course, the same would be true on level ground. A 20mph headwind is also going to load the rear wheel more or less than it's static 40/60 balance if CG and aero center are displaced.

[Mark McM]

If you take something with a steady velocity (stopped on not accelerating - motion is relative) and raise one of its legs (or wheels), the weight shifts toward the lower. That's true whether it is a ladder set on an uphill, a parked car, or a bicycle that has topped out (terminal) at 38 mph on a downhill.

Your model would describe a bike standing on a staircase, but it does not describe a bike on a hill. When on a level platforms, the gravity force is directly opposed by the normal force at the wheel/ground contact point, so the bike remains stationary. As you say, because the CG is tipped forward, weight is shifted to the front wheel.

On a hill, the wheels are on a slope (not level platforms). The gravity force acts in two seperate componenents – one that is normal to the wheel/ground contact point (i.e. pressing the bike against the ground), and second component that is tangential to the slope (acting to accelerate the bike down the slope). Both of these components act through the CG (which is of course why it is called the Center of Gravity). If the bike is allowed to accelerate freely, the weight distribution between the wheels is the same as when on a flat road, because the normal portion of the gravity portion is in the same position and angle with respect to the wheels.

To keep the bike from accelerating down the slope (either stopped or at a constant speed), there must be a force opposing the gravity tangential component. That force can either be from braking (force applied tangentially at the wheel/ground contact point(s)), or wind drag (force applied tangentially approximately at the center of frontal area). In the case of braking, the braking force acts at the wheel/ground contact points, which are offset some distance away from the CG. Because of this offset, a torque is generated on the bike. To balance this out, a counter-torque is generated by shifting weight from the rear wheel to the front wheel.

However, the center of force for air drag is close to the center of frontal area, which isn’t far far from the CG. Therefore, there is little torque generated by the wind drag force. Therefore, where is little need to shift weight between the wheels to counter it. The weight distribution when coasting down a hill at terminal velocity is pretty much the same as when riding on the flat.

So, do this little experiment: Find a set of stairs, and put your bike on a higher step then the rear wheel. Mount the bike in the normal riding position, and note how much of your weight shifts from the saddle to your hands as compared to sitting on the bike when on flat ground. Then, coast down a hill, and see if you can detect any shift in weight from the saddle to the handlebars. I know I sure can’t -even down the steepest hills.


This is why locking up the front brake on a downhill is more dangerous than on a flat - your CG is already tipped toward the front wheel, so it takes less deceleration to fling you over the bars.

There’s a few other things going on here than that. Firstly, when descending a hill, you have the constant acceleration of gravity in addition to your momentum, so you have brake harder just to achieve the same deceleration rate. Secondly, because the gravity force breaks into two components (one normal and one tangential), so the normal force holding you against the road is smaller than when on flat ground, and therefore less braking force is necessary to through you over the bars. There is no forward weight shift before you begin to brake.

You and Madrocket are over thinking this stuff (or underthinking). You know this stuff in the real world, or you'd be falling off ladders and your bikes all the time. Don't want to believe me OR your inner ear? Hang something from your rearview mirror, then drive down and park on a steep hill. You'll find the angle of the dangle is identical whether you're going a steady 35 mph or parked.


And you are underthinking your example. Dangle the same object off your handlebar when riding your bike steady down at a hill at 35 mph. Does it hang at the same angle as when you’re parked on the hill? No – the wind force will be pushing it backward, so it may actually be dangling at an angle backward. This same (wind) force is acting on you. And because the wind is acting so close to your CG, there is little weight shift between the wheels as compared to riding on level ground.

It appears that in your analyses you are neglecting to consider that where force is applied is just as important as its direction and magnitude. If a backwards force (braking or wind drag) is applied below the CG, it shifts weight to the front wheel. But if the backwards force is applied above the CG, it shifts weight to the rear wheel. And if the backward force acts near the CG (like wind drag), there is hardly any weight shift at all. Consider what would happen if you rode your bike fast while holding a large sail high above your head – wouldn’t the air drag from the sail tend to want to rotate you backward, shifting weight to the rear wheel? And wouldn’t this rearward weight shift still happen if you were riding down a hill with the sail above your head?

[zap]

did you get this from my frame order form atmo?

Taking orders again?

[ergott]

OK.......... here's a good one. If you are driving in a car with the windows up and you have a helium balloon on a string and you make a hard right turn what does the balloon do?

a) it swings to the left just like you think it would.

b) it swings to the right just to confuse you.

c) it bobs up and down wildly and then assplodes.

d) it does nothing.



What say you physics experts?

Dave

B

[David Kirk]

B

Cool.

Now what happens if the windows are open.

dave

[ergott]

Cool.

Now what happens if the windows are open.

dave

With all the wind buffeting all around who the hell knows where it's going. You don't have that problem with your Caterham (or should I "say" Katrum).

[Mark McM]

Cool.

Now what happens if the windows are open.

dave


A

(Actually that's a bit misleading. It actually it tries to go in the original direction of the car - it just appears to the occupants like it is swinging to the left.)

[David Kirk]

A

(Actually that's a bit misleading. It actually it tries to go in the original direction of the car - it just appears to the occupants like it is swinging to the left.)

With closed windows the air moves to the outside of the turn and hits the window making a high pressure area which then pushes the balloon toward the inside.

With the window open the air can move to the outside of the turn and rush right out the window and it will try to take the balloon with it.

This is a fun one to try and it works every time.

dave

[jimcav]

The hummingbird will move toward the back of the car. Similarly if you stop suddenly the hummingbird will crash into the window. So will cigarette smoke. A balloon will move toward the back of the car because of blah blah blah air something or other.

the hummingbird is the only bird that can fly backwards--just saying

jim

[MadRocketSci]

what a bunch of geeks

:D

[Mark McM]

the hummingbird is the only bird that can fly backwards--

-- without a headwind.