effect of road grade on VAM
VAM is an easy to calculate proxy for power-to-mass ratio commonly reported for professional riders in European races, where television coverage plus well documented hill profiles facilitates its extraction. VAM is the rate of altitude gained per unit time, for example in meters per hour. I recently recorded 1402 metters per hour in a break-through ride up Old La Honda, breaking my previous PR. The excellent Science of Sport Blog recently posted the following plot of some VAMs from recent Tour history:
VAMs from the Tour de France, from The Science of Sport.
Michele Ferrari's numerical calculation of the effect of road grade on VAM are often used as a reference, by me included, but an analytic form is perhaps more useful.
I'll again assert a simplified, windless bike power-speed model, which is for "PowerTap power" neglecting drivetrain losses, but it is assumed drivetrain losses are a function of total power:
p = fm m s + fw s³,
where p is a fixed power, s is bike speed, and fm and fw are coefficients. This model neglects drivetrain losses, but I'll assume they depend primarily on total power p, and thus are fixed.
Differentiating with respect to speed s yields:
dp / dgrade = ∂p/∂s ∂s/∂grade + ∂p/∂grade
which since dp = 0 yields, keeping only non-zero partial derivatives:
0 = (fm m + 3 fw s²) ∂s/∂grade + m s ∂fm/∂grade
or
∂s/∂grade = ‒ m s ∂fm/∂grade / (fm m + 3 fw s²)
Using a definition for VAM with our linearized grade (sine = tangent):
VAM = grade × s
yields:
( grade / VAM ) (d VAM / d grade) = ( grade / VAM ) [ ∂VAM / ∂grade + (∂VAM / ∂s) (∂s / ∂grade) ]
= 1 ‒ m grade ∂fm/∂grade / (fm m + 3 fw s²).
The quantity on the left is the relative increase in VAM with the relative increase in road grade. For example, what is the ratio of the % increase in VAM with the % increase in grade? This can be simplified by defining δVAM as the fractional increase in VAM and δgrade as the fractional increase in grade.
So given the following:
fm = ( grade + Crr) g
then
∂fm/∂grade = g
where g is the gravitational acceleration, and therefore
δVAM / δgrade = 1 ‒ m g grade / (fm m + 3 fw s²).
If f is the fraction of power coming from wind resistance this yields:
δVAM / δgrade = 1 ‒ ( m g grade s / p ) / ( 1 + 2 f ).
This can be simplified by defining αRR = CRR / (CRR + grade), the fraction of mass-proportional power from wind resistance, yielding
δVAM / δgrade = 1 ‒ (1 ‒ αRR) (1 ‒ f) / (1 + 2 f)
= [ 1 + 2 f ‒ 1 + αRR + f ‒ αRR f) ] / (1 + 2 f)
with a final result:
using the following definitions:
An alternate but equivalent representation would be (I prefer the previous one):
δVAM / δgrade = [ αRR + (3 ‒ αRR) f ] / (1 + 2 f).
Or, if αRR is close to 1, then this more symetric representation may be better, as the function of f on the right is the effect of relative decreases is mass on climbing speed, derived in an earlier post:
δVAM / δgrade = 1 ‒ (1 ‒ αRR) (1 ‒ f) / (1 + 2 f).
Consider the following example:
αRR ≡ 1 / (CRR / grade + 1) = 0.05,
f ≡ fraction of power to wind resistance = 0.12.
Then a 10% relative increase in grade (for example, from 7% to 7.7%) will result in a 3.2% increase in VAM (for example from 1400 m/hr to 1445 m/hr).
There are two limits for this result, each providing a nice check of the derivation. One is where f is essentially zero, in which case it simplifies to αRR. So in this instance the VAM is reduced proportional to the fraction of power going into rolling resistance. In the other limit of αRR → 0, it simplified to 3 f / (1 + 2 f), which represents the decreased power going into wind resistance with steeper grades. There is also the second-order term in the numerator ‒αRR f which has negligible effect on steep relatively smooth roads. To the opposite extreme, where the road is almost flat and therefore f or αRR are close to 1, the result will be close to 1, which implies for very shallow roads dominated by rolling resistance and/or wind resistance a fractional increase in road grade barely affects speed, and so the grade contributes directly to VAM. This is of course correct, so in each of these test limits the formula does the right thing.
VAMs from the Tour de France, from The Science of Sport.
Michele Ferrari's numerical calculation of the effect of road grade on VAM are often used as a reference, by me included, but an analytic form is perhaps more useful.
I'll again assert a simplified, windless bike power-speed model, which is for "PowerTap power" neglecting drivetrain losses, but it is assumed drivetrain losses are a function of total power:
p = fm m s + fw s³,
where p is a fixed power, s is bike speed, and fm and fw are coefficients. This model neglects drivetrain losses, but I'll assume they depend primarily on total power p, and thus are fixed.
Differentiating with respect to speed s yields:
dp / dgrade = ∂p/∂s ∂s/∂grade + ∂p/∂grade
which since dp = 0 yields, keeping only non-zero partial derivatives:
0 = (fm m + 3 fw s²) ∂s/∂grade + m s ∂fm/∂grade
or
∂s/∂grade = ‒ m s ∂fm/∂grade / (fm m + 3 fw s²)
Using a definition for VAM with our linearized grade (sine = tangent):
VAM = grade × s
yields:
( grade / VAM ) (d VAM / d grade) = ( grade / VAM ) [ ∂VAM / ∂grade + (∂VAM / ∂s) (∂s / ∂grade) ]
= 1 ‒ m grade ∂fm/∂grade / (fm m + 3 fw s²).
The quantity on the left is the relative increase in VAM with the relative increase in road grade. For example, what is the ratio of the % increase in VAM with the % increase in grade? This can be simplified by defining δVAM as the fractional increase in VAM and δgrade as the fractional increase in grade.
So given the following:
fm = ( grade + Crr) g
then
∂fm/∂grade = g
where g is the gravitational acceleration, and therefore
δVAM / δgrade = 1 ‒ m g grade / (fm m + 3 fw s²).
If f is the fraction of power coming from wind resistance this yields:
δVAM / δgrade = 1 ‒ ( m g grade s / p ) / ( 1 + 2 f ).
This can be simplified by defining αRR = CRR / (CRR + grade), the fraction of mass-proportional power from wind resistance, yielding
δVAM / δgrade = 1 ‒ (1 ‒ αRR) (1 ‒ f) / (1 + 2 f)
= [ 1 + 2 f ‒ 1 + αRR + f ‒ αRR f) ] / (1 + 2 f)
with a final result:
| δVAM / δgrade = [ (1 ‒ f) αRR + 3 f ] / (1 + 2 f) |
using the following definitions:
| δVAM | fractional increase in VAM |
| δgrade | fractional increase in road grade |
| αRR | fraction of mass-proportional power from rolling resistance |
| f | fraction of total power from wind resistance |
An alternate but equivalent representation would be (I prefer the previous one):
δVAM / δgrade = [ αRR + (3 ‒ αRR) f ] / (1 + 2 f).
Or, if αRR is close to 1, then this more symetric representation may be better, as the function of f on the right is the effect of relative decreases is mass on climbing speed, derived in an earlier post:
δVAM / δgrade = 1 ‒ (1 ‒ αRR) (1 ‒ f) / (1 + 2 f).
Consider the following example:
αRR ≡ 1 / (CRR / grade + 1) = 0.05,
f ≡ fraction of power to wind resistance = 0.12.
Then a 10% relative increase in grade (for example, from 7% to 7.7%) will result in a 3.2% increase in VAM (for example from 1400 m/hr to 1445 m/hr).
There are two limits for this result, each providing a nice check of the derivation. One is where f is essentially zero, in which case it simplifies to αRR. So in this instance the VAM is reduced proportional to the fraction of power going into rolling resistance. In the other limit of αRR → 0, it simplified to 3 f / (1 + 2 f), which represents the decreased power going into wind resistance with steeper grades. There is also the second-order term in the numerator ‒αRR f which has negligible effect on steep relatively smooth roads. To the opposite extreme, where the road is almost flat and therefore f or αRR are close to 1, the result will be close to 1, which implies for very shallow roads dominated by rolling resistance and/or wind resistance a fractional increase in road grade barely affects speed, and so the grade contributes directly to VAM. This is of course correct, so in each of these test limits the formula does the right thing.
Comments
The effect of wind on a uphill was exactly something I was hunting for.
Please critique my article on climbing rate if you can. http://cozybeehive.blogspot.com/2009/07/rate-of-climbing-uphill-explained.html
Its lengthy but I hope I have beaten it to death in a fashion people can understand. I like your analytic treatment of it, but I have also done a visual and vector treatment of it, that lets people actually see what with their eyes what the climbing rate or vertical speed is really.
However, some of the things I have left out is how climbing rate depends on altitude and what not. I think your work here and Ferrari's field work at 53x12 should fill in for that absence.