Road percent grade and power

[BtonRider]

I came across this statement online "a 5% grade requires a forward force equal to 5% of the weight of the object (above and beyond the force it takes to overcome surface resistance on flat ground at the same speed)." Has anyone considered this?

I plugged a few numbers into Analytic cycling and I'm not sure if this statement is correct. It looks like it takes 58% more power to maintain 8m/s (for a 75kg person) on a 1% grade than a flat road. However, it takes only 5% more power to maintain that speed on a 20% road vs a 19% road. Of course I don't think too many people can put out enough power to do 8m/s on a 20% road (1278W) for any extended period. Now that all being said, the power difference (in Watts) betweeen a 1% and 2% is the same as between 19% and 20% (in the case of the example I've given 58.8W), so to me that disagrees with what the above author said.

I was just wondering if anyone has a good explanation for the relationship of percent road grade to power output.

 

[daveryanwyoming]

I came across this statement online "a 5% grade requires a forward force equal to 5% of the weight of the object (above and beyond the force it takes to overcome surface resistance on flat ground at the same speed)." Has anyone considered this?

I haven't bothered to validate or challenge this statement, but note that it talks about change in force, not change in power. Power is equal to force * velocity so the two aren't interchangeble.

...I was just wondering if anyone has a good explanation for the relationship of percent road grade to power output.

Sure, just go to this page: http://www.analyticcycling.com/ForcesPower_Page.html

The key thing is this section:
[size=-1]Wind Resistance[/size] [size=-1]Fw = 1/2 A Cw Rho Vmps2[/size] [size=-1]
Rolling Resistance[/size] [size=-1]Frl = Wkg 9.8 Crr[/size]
[size=-1]Gravity Forces[/size] [size=-1]Fsl = Wkg 9.8 GradHill[/size] [size=-1]
Power[/size] [size=-1]RiderPower = (Fw + Frl + Fsl) Vmps

[/size] The last line gives you a very good estimate of total power output necessary to maintain a given speed. The first three lines break down those forces which change based on things like aerodynamics, speed, tire rolling resistance and pavement type as well as the grade you were asking about. At slower speeds on a hill the gravity forces will dominate but you still get some rolling resistance and wind resistance so it's not quite a linear relationship to grade especially since your frontal area tends to increase as you sit up to climb.

Anyway, the answer is in those formulas, take the time to plug the first three into the fourth and you'll see what happens as you keep speed (and Crr, CdA, etc.) constant and increase grade by a certain percent.

-Dave

 

[frenchyge]

I came across this statement online "a 5% grade requires a forward force equal to 5% of the weight of the object (above and beyond the force it takes to overcome surface resistance on flat ground at the same speed)."

That's not *exactly* correct, but is close enough for road work and very small angles.

 

[BtonRider]

I haven't bothered to validate or challenge this statement, but note that it talks about change in force, not change in power. Power is equal to force * velocity so the two aren't interchangeble.

Okay, I've just tried to simplify this. For example, if you assume a constant velocity on two different grades, the cross section, rolling resistance, etc should be constant as well. The question becomes how much more power do you need to put out to maintain a given speed when the road goes from flat to 5%?

Based on the equation you mentioned (power equals force times velocity), if we keep v constant can't we say

P1=F1*V (on a flat road for example)
P2=F2*V (on a 5% grade)

If F2=1.05*F1
then P2=1.05*F1*V
and P2=1.05*P1
(so a 5% grade should require only 5% more power than a flat road to maintain a given velocity)

This doesn't agree with what analytic cycling comes up with though. For me to maintain 8m/s (17.9mph) on that rise my power output has to increase 388%

 

[daveryanwyoming]

...The question becomes how much more power do you need to put out to maintain a given speed when the road goes from flat to 5%?...

O.K., I tried to avoid the grunt work, but here it goes:

power = V*Fw + V*Frl + v*Fsl
= V*Fw + V*Frl + V*W*g*Grad

Holding everything constant except Grad (not realistic if you sit up on a hill) and taking the first derivative with respect to Grad

dpower/dGrad = 0 + 0 + V*W*g

Using the analyticcycling defaults of 8 m/s, 75kg and g = 9.8 m/s^2

V*W*g = 5880
That constant represents delta pwr/delta Grad, multiply it by 1% to break it down to power per percent change in Grade.

V*W*g*0.01 = 58.8 watts/1% change in grade.

So for a 75 kg rider(complete weight with bike and kit) riding a steady 8 m/s each additional 1% of steepness in grade requires an additional 58.8 watts. That works out to an additional 396 watts for a 5% grade and ~590 watts for a 10% grade.

Hope that makes sense.
-Dave

 

[BtonRider]

So for a 75 kg rider(complete weight with bike and kit) riding a steady 8 m/s each additional 1% of steepness in grade requires an additional 58.8 watts. That works out to an additional 396 watts for a 5% grade and ~590 watts for a 10% grade.

I get the same results when I use Analytic Cycling.

So, what I draw from all of this is the additional amount of power necessary to maintain a given speed when the grade changes from 19% to 20% is the same additional power needed when the grade changes from flat to 1%. This makes intuitive sense to me, and I'm glad the math supports this.

Then, how does the percent grade relate to the additional power needed to go from flat to 1%? I did the same calculation and found a 58.8W increase (given the default settings on analytic cycling); but how does that 58.8W relate to 1%? I mean 58.8W is a power increase of 50%.

 

[frenchyge]

Then, how does the percent grade relate to the additional power needed to go from flat to 1%?

Let's look at the initial statement:

"a 5% grade requires a forward force equal to 5% of the weight of the object (above and beyond the force it takes to overcome surface resistance on flat ground at the same speed)."

The bolded portion means that the flat riding resistance needs to first be *subtracted* from the 5% grade resistance to find the linearly dependent uphill resistance. That's not the same as dividing the 5% grade resistance by 1.05 to get the flat resistance.

If you want to see the thumbrule in action, it would be best to compare two uphill scenarios rather than an uphill scenario vs. a flat scenario. In the latter comparison, when you subtract out the flat resistance from both scenarios, you're left trying to solve for a ratio where one of the terms is zero, and that doesn't work well.

I did the same calculation and found a 58.8W increase (given the default settings on analytic cycling); but how does that 58.8W relate to 1%? I mean 58.8W is a power increase of 50%.

Forget about the flat riding scenario, since the conditions of the stated thumbrule don't apply to riding on the flats. If each 1% increase in grade adds 58.8w to the power requirement, then that means a fixed 1% of the rider's weight is being added to the resistance (which is then multiplied by the constant velocity to get the fixed power increment).

 

[BtonRider]

Forget about the flat riding scenario, since the conditions of the stated thumbrule don't apply to riding on the flats. If each 1% increase in grade adds 58.8w to the power requirement, then that means a fixed 1% of the rider's weight is being added to the resistance (which is then multiplied by the constant velocity to get the fixed power increment).

Okay that makes sense... and works out mathematically. Thanks.

Although I think climbing a 20% hill feels like I'm hauling more than 20% more weight (more like 200%)

It just seems like there should be some way of relating/comparing 1% to 0%.

 

[daveryanwyoming]

.. Then, how does the percent grade relate to the additional power needed to go from flat to 1%? .... I mean 58.8W is a power increase of 50%.

You can't think of the change from flat to even the slightest uphill as a percentage increase. It's the addition of an entirely new loss term. When riding on the flat you're not lifting your weight at all against gravity. Move to a 1% grade and suddenly you have a new way that power gets sucked out of the system, you need all the power you were previously generating on the flats to overcome air and rolling resistance but now you've got a new enemy in the form of gravity.

So it doesn't make any sense to think of the gravity loss in terms of a percentage of the rolling + air losses. They're different beasts with some common components (like weight and velocity) but one is not a percentage of the other.

Frenchyge and I are saying basically the same thing, you've got to subtract the power terms related to rolling resistance and air resistance and then consider the gravity term by itself.

-Dave

 

[daveryanwyoming]

...It just seems like there should be some way of relating/comparing 1% to 0%.

There is, it's just additive, not multiplicative.

 

[frenchyge]

Although I think climbing a 20% hill feels like I'm hauling more than 20% more weight (more like 200%)

You're hauling 20% of your body weight, which is 20x (2000%) more hauling weight than when you're climbing a 1% grade.

It just seems like there should be some way of relating/comparing 1% to 0%.

If you can express 1% as a percentage of zero (or vice versa), then you'll have found your relationship.

 

[sugaken]

If you can express 1% as a percentage of zero (or vice versa)...

The vice versa part is easy, isn't it?; zero is 0% of 1%

Ken

 

[531Aussie]

try the Kreuzotter calculator, and see if it's any different

http://www.kreuzotter.de/english/espeed.htm