What is the truth behind bike weight? Does it really help THAT much?



TheDarkLord

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sogood said:
I am no physicist and there are lots of far better qualified people here to answer this.

My understanding is that given the same bike apart from weight (same resistive forces), the propulsive force on the heavier bike would be greater due to its mass (g is a constant and same for both bikes). So as a result, the net force (propulsive) on the heavier bike is greater than the lighter bike).

Am I wrong to explain it this way? :eek:
The force is higher. But acceleration is force per unit mass. So, the acceleration by gravity alone will be the same for all bikes irrespective of their mass. It is the drag force that makes all the difference. I'm not sure if that equation has an analytic solution...
 

TheDarkLord

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TheDarkLord said:
The force is higher. But acceleration is force per unit mass. So, the acceleration by gravity alone will be the same for all bikes irrespective of their mass. It is the drag force that makes all the difference. I'm not sure if that equation has an analytic solution...
Ok, I just looked up some notes on differential equations. If the resistance is proportional to velocity, then there is a very easy analytic solution. Long story short, if the drag force is k*v (where k is a constant that depends on the bike+rider parameters), then the terminal velocity will be mg/k (g = acceleration due to gravity). So, a 2% heavier bike should have a 2% higher terminal velocity. If on the other hand, the drag force is proportional to the square of velocity, then I don't know yet how to solve that differential equation...
 

sogood

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TheDarkLord said:
The force is higher. But acceleration is force per unit mass. So, the acceleration by gravity alone will be the same for all bikes irrespective of their mass. It is the drag force that makes all the difference. I'm not sure if that equation has an analytic solution...
But, but... F = ma, or mg in this case. So the determinant of that force is the variable mass. Why do I get a feeling that you don't need to bring calculus into this?
 

dhk2

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sogood said:
But, but... F = ma, or mg in this case. So the determinant of that force is the variable mass. Why do I get a feeling that you don't need to bring calculus into this?
Agree calculus isn't needed if the mass and force (ie, gradient) is constant over time and we just want to look at steady-state conditions; ie, top speed. Just the opposite of climbing up, the rate of potential energy lost on the descent would be the same as the power "generated" from the hill (neglecting rolling resistance), so that watts "generated" on the descent would be expressed as V (m/sec) x % slope x force (nts).

The other side of the equation (at top speed) is power required for aero drag. Rather than messing with the aero drag equation, which is a function of air density, drag coefficient, frontal area and velocity cubed, then solving for V by taking a square root, there are calculator spreadsheet programs that will estimate this for us.

Plugged some rider weights into a spreadsheet program for a 10% downhill grade. Using the cda figure for "racers crouch", the program says a 155 lb (70 kg) rider on a 20 lb bike will hit a top speed of 40 mph, while a 200 lb (91 kg) rider will go almost 45 mph. These results line up well with my experience: I weigh 85 kg, and find I often pass the lighter climbers on the descents.
 

JohnO

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All the same, I'd rather put forth a bit more energy on a descent to make up for less weight, than a lot more energy on a climb to make up for more weight.

That goes mainly for myself. It was a lot easier for me to drop ten pounds than my bike to drop ten pounds.

And on a descent, most of the advantage of more weight can be overcome with drafting. The same isn't nearly as true for climbing.
 

Crankyfeet

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The heavier rider + bike goes faster in descent (if you take pedalling out of the equation).

The principle is the same reason why a ball-bearing hits the ground if dropped from a tall building, before a same-sized plastic ball full of air (in the atmosphere, not in a vacuum).
 

Crankyfeet

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JohnO said:
All the same, I'd rather put forth a bit more energy on a descent to make up for less weight, than a lot more energy on a climb to make up for more weight.

That goes mainly for myself. It was a lot easier for me to drop ten pounds than my bike to drop ten pounds.

And on a descent, most of the advantage of more weight can be overcome with drafting. The same isn't nearly as true for climbing.
Also, on descents, there is often a maximum speed anyway, so if people are braking into turns, the effect of momentum is nullified somewhat, especially if the lighter rider can take the turn faster.

There is no comparitive restriction on the lighter rider on the uphill though. He doesn't have to brake more than the slower heavier rider (assuming the riders are delivering the same absolute power to the pedals).

I trained with a 100 pound lady recently (Eden from here) who almost had to keep herself in zone 5 just to keep up with me on a long descent, when I was just coasting on the pedals. I could even keep up with her easily by not pedalling when I went down behind her (I stayed out of her draft). We were doing hill reps. She said that her having to do more work on descents in races was sometimes a big disadvantage to her as she was sometimes at a high heart rate already before the next climb started. She had the advantage going uphill though.
 

Tapeworm

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Crankyfeet said:
The heavier rider + bike goes faster in descent (if you take pedalling out of the equation).

The principle is the same reason why a ball-bearing hits the ground if dropped from a tall building, before a same-sized plastic ball full of air (in the atmosphere, not in a vacuum).

But what would hit first, a ball-bearing or a slightly bigger ball-bearing?

A heavy bike is only faster if its aerodynamics are equal or better than a lighter bike?

A light bike that had better aerodynamics would be as fast or faster than the heavier bike?

If two riders being the same weight and build but riding bikes with say a 1kg difference, how much difference are we talking? Greater difference depending on the slope?
 

sogood

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Tapeworm said:
A heavy bike is only faster if its aerodynamics are equal or better than a lighter bike?
Yes.
A light bike that had better aerodynamics would be as fast or faster than the heavier bike?
Depends on how much more aero. Specific numbers starts to matter.
If two riders being the same weight and build but riding bikes with say a 1kg difference, how much difference are we talking? Greater difference depending on the slope?
Put the numbers into an appropriate equation, or go and test it with the real thing. And if you are racing for $100k prize money, then every bit counts. Otherwise, just sit back and enjoy the ride.
 

TheDarkLord

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sogood said:
But, but... F = ma, or mg in this case. So the determinant of that force is the variable mass. Why do I get a feeling that you don't need to bring calculus into this?
What matters for the speed - force or acceleration? Sure, the force is larger, but that force is acting on a larger mass. The acceleration (not counting drag) is the same. Yes, you don't need calculus if you neglect drag. But you need calculus if you introduce drag, since drag depends on velocity, and so you have velocity and a derivative of velocity in the same equation.

Edited to add: dhk2's post clarifies this.
 

dhk2

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Tapeworm said:
But what would hit first, a ball-bearing or a slightly bigger ball-bearing?

A heavy bike is only faster if its aerodynamics are equal or better than a lighter bike?

A light bike that had better aerodynamics would be as fast or faster than the heavier bike?

If two riders being the same weight and build but riding bikes with say a 1kg difference, how much difference are we talking? Greater difference depending on the slope?
Good questions. Let me take a shot at some quick answers:

1. If the slightly bigger ball bearing was made of the same density material, with the same surface finish, it would reach a higher terminal velocity and hit sooner. This is because of the properties of the equations for weight and air resistance. The weight driving the ball down is due to the volume of the sphere which relates to the cube of the diameter. However, the air drag is a function of the frontal area, which is a function of the square. These same equations are why the heavier rider has the advantage downhill over the lighter one, assuming he's in the same relative aero position.

2. To a point, yes. But it's the total weight and drag of the bike and rider that determine top speed, and the bike really doesn't matter much since we riders have much more of each. Aero positioning on the bike is a huge factor (sitting up vs down in TT position with elbows and knees tight. I've read that we can gain more aero drag reduction from wearing an aero helmet, taking off the gloves and wearing a tight skinsuit than a wheelset can provide.

3. Same answer; weight and aero of the rider is more important.

4. Using the calculator spreadsheet, looks like 1 kg more (on bike or rider's body) increases speed on a 10% slope from 40.1 to 40.3 mph. On a 5% downgrade, the numbers are 31.1 and 31.2 mph. These figures used "hands on drops, elbows locked" drag estimates. Substituting the cdA value for "full racing tuck" takes the speed up to 36.9 mph...which is huge. Most of us have experienced this on a long descent: tuck in well, and a few seconds later you're in the passing zone.
 

TheDarkLord

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dhk2 said:
4. Using the calculator spreadsheet, looks like 1 kg more (on bike or rider's body) increases speed on a 10% slope from 40.1 to 40.3 mph. On a 5% downgrade, the numbers are 31.1 and 31.2 mph. These figures used "hands on drops, elbows locked" drag estimates. Substituting the cdA value for "full racing tuck" takes the speed up to 36.9 mph...which is huge. Most of us have experienced this on a long descent: tuck in well, and a few seconds later you're in the passing zone.
What is the mass of the bike + rider that you assumed? Just curious on what the percentage decrease in mass is.
 

Crankyfeet

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One interesting factor on climbing and descents (though it doesn't have anything to do with bike weight) is that quite often the heavier rider has more power (is stronger) on the climbs to compensate for his heavier weight. The rule of thumb - power to weight ratio - is the same for a 20% heavier rider if his power output is 20% higher (this is not an exact indicator as relative speed will also be a function of the steepness of the slope).

On the descents however, the heavier rider not only can build up more momentum and have a higher terminal velocity (assuming drag effects in a tuck position are roughly the same) but the heavier rider, in the above example, also has 20% more available power than the lighter rider if pedalling to increase velocity. It can be a double whammy effect.

The physics advantage of having a lighter weight on the climbs is still more significant however... hence why the best climbers are usually lighter in build.
 

Eden

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Crankyfeet said:
On the descents however, the heavier rider not only can build up more momentum and have a higher terminal velocity (assuming drag effects in a tuck position are roughly the same) but the heavier rider, in the above example, also has 20% more available power than the lighter rider if pedalling to increase velocity. It can be a double whammy effect.

I bother with any physics, but I can tell you from experience that this is way too true. At only a little over 100 lbs its so easy to get dropped going down hill that its not funny... I can redline chasing going down, which is not so useful when it levels out or starts going up again.. AND I believe that at what I weigh its is actually more difficult to corner (heavier folks stick to the road better - at least my experience riding around with a 15lb load in my messenger bag and having it feel way better cornering - the opposite of what one might expect makes me believe this to be true) and I get bounced around more on rough roads which also can make descents spookier..
 

TheDarkLord

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Eden said:
I bother with any physics, but I can tell you from experience that this is way too true. At only a little over 100 lbs its so easy to get dropped going down hill that its not funny... I can redline chasing going down, which is not so useful when it levels out or starts going up again.. AND I believe that at what I weigh its is actually more difficult to corner (heavier folks stick to the road better - at least my experience riding around with a 15lb load in my messenger bag and having it feel way better cornering - the opposite of what one might expect makes me believe this to be true) and I get bounced around more on rough roads which also can make descents spookier..
I guess cornering is related to rolling friction. For the same coefficient of rolling friction, a heavier bike+rider combo will have larger rolling friction, and will probably contribute to better cornering. But you can probably control this to some extent (plus how much you are bounced around) through the pressure to which the tires are inflated.
 

Crankyfeet

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TheDarkLord said:
I guess cornering is related to rolling friction. For the same coefficient of rolling friction, a heavier bike+rider combo will have larger rolling friction, and will probably contribute to better cornering. But you can probably control this to some extent (plus how much you are bounced around) through the pressure to which the tires are inflated.
Yeah... but the larger friction is compensated by the larger component lateral force that a heavier rider has... doesn't it? You know... centrifugal force (or is it centripetal??? I always get confused on those two terms)
 

TheDarkLord

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Crankyfeet said:
Yeah... but the larger friction is compensated by the larger component lateral force that a heavier rider has... doesn't it? You know... centrifugal force (or is it centripetal??? I always get confused on those two terms)
Yeah, that's true. To clarify, the force felt by the rider during a curve is centrifugal force, while the actual acceleration itself (due to change in velocity) is directed in the opposite direction and is called centripetal acceleration. So, I guess the coefficient of friction is the primary factor when it comes to cornering without the tire washing out. So, why would a heavier rider be able to corner better?
confused0006.gif
 

dhk2

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TheDarkLord said:
What is the mass of the bike + rider that you assumed? Just curious on what the percentage decrease in mass is.
I used 155 lb rider with 20 lbs of bike and equipment, then added 2 lbs....not exactly a kg, but then all these numbers are just approximate due to the input variables not being exactly known. Anyway, looks like a 1.3% increase in weight yields about a 0.6% increase in speed. This is expected, since the equation balances force (weight) against the square of the speed. (If you square 1.0064 you'll get 1.013).

Actually, that's the reason us heavier guys don't have an advantage in the hills. For the same power, that extra kg we drag up that steep hill costs a straight % of our total weight, while going down, we only get the square root of that % back.
 

TheDarkLord

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dhk2 said:
I used 155 lb rider with 20 lbs of bike and equipment, then added 2 lbs....not exactly a kg, but then all these numbers are just approximate due to the input variables not being exactly known. Anyway, looks like a 1.3% increase in weight yields about a 0.6% increase in speed. This is expected, since the equation balances force (weight) against the square of the speed. (If you square 1.0064 you'll get 1.013).

Actually, that's the reason us heavier guys don't have an advantage in the hills. For the same power, that extra kg we drag up that steep hill costs a straight % of our total weight, while going down, we only get the square root of that % back.
Yes, it agrees with theory. I was gonna post this before, but was waiting for your input regarding the parameters you used... I posted yesterday that for a drag force that is dependent on velocity, the terminal velocity is proportional to mass. Wikipedia has an approximate analytic solution for the case where the drag is proportional to velocity square (involving hyperbolic functions; you can see it here), and as you say, the terminal velocity is proportional to the square root of mass. So, a 1% reduction in weight will give rise to 0.5% reduction in terminal velocity.

But is it really true that the power output of a heavier rider is proportional to the weight (Cranky's post)? Don't see why climbers like Chicken would be so lean then...
 

Crankyfeet

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TheDarkLord said:
So, why would a heavier rider be able to corner better?
confused0006.gif
Maybe because of the effects that Eden mentioned. The lighter rider is getting a more wavering frictional effect due to bumps and undulations, which bounce the lighter rider more.

The proverbial chain (in this case - grip on the road throughout a corner) is only as strong as the weakest link. Once slipping starts, the force required is much less to sustain slipping (less than the force required to initially cause it). It is the same theory behind anti-lock brake systems in cars (which is - prevent skidding => more braking force).
 

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