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Physics of Descending

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Old 03-22-15 | 08:47 PM
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Originally Posted by Pirkaus
Frontal area to weight is inversely proportional.
Cycling: Uphill and Downhill
My knee-jerk estimate is frontal area proportional to mass to 2/3 power. More or less.
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Old 03-22-15 | 08:54 PM
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Originally Posted by wphamilton
My knee-jerk estimate is frontal area proportional to mass to 2/3 power. More or less.
Yeah, just about.
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Old 03-22-15 | 09:01 PM
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Originally Posted by Pirkaus
When descending I will pedal until I spin out my legs, then tuck in tight.
Sorry for the noob question, but is this common practice? I noticed today while riding that on descents at high speed I eventually ran out of resistance in the highest gear and started spinning out, so I just did as you said above. What speed does that typically happen to you at?
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Old 03-22-15 | 09:04 PM
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Originally Posted by chilidawgnv
Sorry for the noob question, but is this common practice? I noticed today while riding that on descents at high speed I eventually ran out of resistance in the highest gear and started spinning out, so I just did as you said above. What speed does that typically happen to you at?
Yep, spin up and then rest. Use the rest time to ponder changing to a 53 tooth crank.
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Old 03-22-15 | 09:15 PM
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Originally Posted by znomit
Yep, spin up and then rest. Use the rest time to ponder changing to a 53 tooth crank.

'Preciate it. I'll jot that down on my growing list of "to upgrade."
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Old 03-22-15 | 10:06 PM
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I read that at some point on a high speed descent, the energy that it would to take to increase your speed becomes so great, and the gains so little, that it makes more sense to become aero and spend the time recovering.
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Old 03-22-15 | 10:38 PM
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Originally Posted by colnago62
I read that at some point on a high speed descent, the energy that it would to take to increase your speed becomes so great, and the gains so little, that it makes more sense to become aero and spend the time recovering.
Above 60kph and you're getting close to 1000W from gravity. Adding an extra 400W will give you a little speed but it's much better to find anyone to tuck behind. Even if you're not right on their wheel there are significant benefits to be had from drafting.

Without a technical descent it's a poor place to try and distance your competitors. Like attacking into a headwind.
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Old 03-22-15 | 11:36 PM
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Old 03-23-15 | 06:14 AM
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Originally Posted by gregf83
Above 60kph and you're getting close to 1000W from gravity. Adding an extra 400W will give you a little speed but it's much better to find anyone to tuck behind. Even if you're not right on their wheel there are significant benefits to be had from drafting.

Without a technical descent it's a poor place to try and distance your competitors. Like attacking into a headwind.
Actually headwinds and upslopes are the preferred places to "try and distance your competitors". You don't go off the front on a break when it is easy for folks to catch you. That is especially true when a sharp angle turn is ahead that changes the wind or slope from against you to with you. If you can make it to the turn way ahead of the pack, you are golden.
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Old 03-23-15 | 06:18 AM
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Originally Posted by asgelle
How is ignoring rolling resistance splitting hairs? There are two terms in the equation of motion that scale with mass: 1) the driving force due to gravity, and 2) the ******ing force due to rolling resistance. How can you justify including the first, but ignore the second? Perhaps you meant changes in the coefficient of rolling resistance which will depend on tire size, construction, and inflation pressure; but not on weight.
Sure, anyone with flat knobbies and grit-filled hubs is going to have a hard time keeping up with anyone on a slick bike,
but in this case who's to say which rider has equipment with more or less rolling resistance? For present discussion purposes we can assume they are the same for each, can't we?
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Old 03-23-15 | 06:21 AM
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Originally Posted by datlas
But on today's ride, I noticed that my fellow riders were really out-descending me. By a few MPH.

It appears that, if you assume negligble rolling resistance and that coefficient of drag are the same for each rider (which is a leap, bigger riders are likely to have more frontal area and thus more drag), that terminal velocity while going downhill is directly proportional to the square root of the mass of rider + bike.

Therefore, assuming that my fellow riders + bike weight are 20% more than me (which I think is accurate since my weight + bike is around 170 and theirs is around 205, which is 20% more than mine), their velocity going down a hill is roughly 10% faster than mine (square root of 1.2 is 1.095).

Well it's either all that, or maybe they just had ceramic bearings?
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Old 03-23-15 | 08:21 AM
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Rolling resistance increases linearly with speed, aero drag is quadratic. And heavier riders do not have proportionally higher resistance, assuming they have properly sized and inflated tires.
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Old 03-23-15 | 08:29 AM
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Originally Posted by gregf83
Without a technical descent it's a poor place to try and distance your competitors.
I got the impression from the OP that he wasn't describing a hypothetical but an empirical (and routine) situation...and unless he lives in some wonderful mythological land where all descents are straight downhill dragstrips, they're all "technical" to some extent ...which is a long way of saying, if OP is getting dropped on descents it probably has more to do with cycling technique than physics.
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Old 03-23-15 | 09:00 AM
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Originally Posted by Bob Ross
I got the impression from the OP that he wasn't describing a hypothetical but an empirical (and routine) situation...and unless he lives in some wonderful mythological land where all descents are straight downhill dragstrips, they're all "technical" to some extent ...which is a long way of saying, if OP is getting dropped on descents it probably has more to do with cycling technique than physics.
I AM getting dropped on descents and I do believe my weight is the cause. I don't think it's technique per se unless you consider getting more aero part of "technique." I am not comfortable tucking my butt under the saddle nose and putting my face on my stem, so unless I gain more weight I will have to work harder to stay in the big boys' slipstream on the descents.
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Old 03-23-15 | 09:23 AM
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Originally Posted by datlas
I AM getting dropped on descents and I do believe my weight is the cause. I don't think it's technique per se unless you consider getting more aero part of "technique." I am not comfortable tucking my butt under the saddle nose and putting my face on my stem, so unless I gain more weight I will have to work harder to stay in the big boys' slipstream on the descents.
Getting more aero and cornering better are what you do. I'm not a heavy rider, usually 148-158 lbs. but am a somewhat famous local descender. Heavy people can't get near me. I think it's partly my build (I have short legs) and partly because I used to ski downhill and so have a good feel for aero and don't mind speed. It's all about getting tiny, thus reducing your swept area, and about getting rid of stuff that catches the wind, like your arms, legs, chest, etc. Rolling resistance matters. Bike shape doesn't seem to matter much. Wheels maybe a little, but nothing compared to position and cornering ability. Hubs don't matter at all. Weight doesn't seem to matter much, I think because heavier people have a hard time getting rid of drag and maybe as was mentioned above, they accelerate more slowly. I also stop pedaling sooner than many people. Pretty hard to overcome the drag of your legs being out in the wind and having legs pop your chest higher. On the usual "spin-out" grade, I'll stop pedaling at around 35 mph.
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Old 03-23-15 | 10:06 AM
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Originally Posted by gsa103
Rolling resistance increases linearly with speed, aero drag is quadratic.
That is true (but irrelevant)

Originally Posted by gsa103
And heavier riders do not have proportionally higher resistance, assuming they have properly sized and inflated tires.
That is false. Rolling resistance is the coefficient of rolling resistance times the normal force on the surface which is proportional to weight. https://www.recumbents.com/WISIL/Mart...%20cycling.pdf Eq. 5. Of course, it is possible that through changing tire size, construction, and pressure, a heavier rider has a Crr low enough that he can overcome the increase in normal force compared to a lighter one, but we are considering riders on equal equipment.
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Old 03-23-15 | 10:08 AM
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Originally Posted by kbarch
Sure, anyone with flat knobbies and grit-filled hubs is going to have a hard time keeping up with anyone on a slick bike,
but in this case who's to say which rider has equipment with more or less rolling resistance? For present discussion purposes we can assume they are the same for each, can't we?
We can assume they have the same coefficient of rolling resistance (after all that is what is equipment dependent). But for equal equipment, the heavier rider will have higher rolling resistance than the lighter one.
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Old 03-23-15 | 10:13 AM
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Originally Posted by wphamilton
My knee-jerk estimate is frontal area proportional to mass to 2/3 power. More or less.



Originally Posted by rpenmanparker
Yeah, just about.
So the air resistance would vary with (strike that, sorry)

Last edited by wphamilton; 03-23-15 at 01:24 PM. Reason: Duh
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Old 03-23-15 | 10:37 AM
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Originally Posted by chilidawgnv
'Preciate it. I'll jot that down on my growing list of "to upgrade."

Until you spin out 53/11 and you see those 55t aero chainrings
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Old 03-23-15 | 10:45 AM
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Two balls are dropped from the top of a building, one is heavier the other one is lighter, they hit the ground at exactly the same time. If you change the shape of one ball so it has more frontal area, thus more drag, it will take longer to reach the bottom. Therefore, gravitational acceleration is independent of mass. If a heavier rider and a light rider can have the same coefficient of drag, they will descent at the same speed.

Edit: until one reaches terminal velocity first, which is the light rider.

Last edited by greenlight149; 03-23-15 at 10:54 AM.
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Old 03-23-15 | 10:49 AM
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Originally Posted by asgelle
We're not looking at the effect. but the change in the effect with changes in mass. A 20% change in mass increases the driving force from gravity by 20% and also increases the ******ing force from rolling resistance by 20%. Since the OP wants to neglect this change, I'd say it's incumbent on him to show it's negligible. From the above, it's not obvious that it must be.
120% of small is still small. Rolling resistance is also simply proportional to speed whereas air resistance is a squared relationship. It's pretty clear to me that it's a negligible effect when talking about a 30-40mph descent. If you've ever ridden on large diameter rollers, you know it doesn't take much power to push wheel speeds well higher than 30mph, and a smooth road is the equivalent of a very very very large diameter roller.

Whatever the effect of rolling resistance, I think it is pretty clear that aerodynamics is the much more dominant effect for this problem. Rolling resistance and frontal areas (assuming equivalent tucks) are both second order effects when talking about descending speed.
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Old 03-23-15 | 10:58 AM
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Originally Posted by rpenmanparker
Actually headwinds and upslopes are the preferred places to "try and distance your competitors". You don't go off the front on a break when it is easy for folks to catch you. That is especially true when a sharp angle turn is ahead that changes the wind or slope from against you to with you. If you can make it to the turn way ahead of the pack, you are golden.
I agree on the upslopes obviously but not into a a headwind. It's much easier for a pack to catch you in a headwind than a tailwind so if you have a choice it's better to attack with a tailwind.
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Old 03-23-15 | 11:02 AM
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Originally Posted by greenlight149
Two balls are dropped from the top of a building, one is heavier the other one is lighter, they hit the ground at exactly the same time. If you change the shape of one ball so it has more frontal area, thus more drag, it will take longer to reach the bottom. Therefore, gravitational acceleration is independent of mass. If a heavier rider and a light rider can have the same coefficient of drag, they will descent at the same speed.
incorrect. Do the math. We'll wait.

Without drag, the light and heavy objects will accelerate due to gravity at the same rate because gravitational force is acting unopposed. With drag, the heavier object accelerates faster than the lighter one even assuming equivalent coefficients of drag. Gravitational force acts on the higher mass of heavy objects more than light objects while drag is proportional (squared, in the case of aero drag) to velocity and is not dependent on mass. Thus, the same drag force acting on a heavy vs. light object creates a resultant force that is larger for the heavier object and smaller for the lighter object, causing the heavier object to accelerate faster.
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Old 03-23-15 | 11:20 AM
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Originally Posted by Brian Ratliff
incorrect. Do the math. We'll wait.

Without drag, the light and heavy objects will accelerate due to gravity at the same rate because gravitational force is acting unopposed. With drag, the heavier object accelerates faster than the lighter one even assuming equivalent coefficients of drag. Gravitational force acts on the higher mass of heavy objects more than light objects while drag is proportional (squared, in the case of aero drag) to velocity and is not dependent on mass. Thus, the same drag force acting on a heavy vs. light object creates a resultant force that is larger for the heavier object and smaller for the lighter object, causing the heavier object to accelerate faster.
Incorrect

Acceleration of two objects of different mass is identical, given the same coefficient of drag.

A=f/m
F=mg
A=mg/m=g independent of mass

You are thinking about terminal velocity, which is in favor of the heavier rider. Which I mentioned in my post, the light rider will reach their terminal velocity first, which is lower than the heavy rider.

In terms of acceleration with same coefficient of drag, they will be identical.
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Old 03-23-15 | 11:22 AM
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Do the math. We'll wait.
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