DirtyWorks said:
Eshnar,
All good questions. I'm looking for what percentage of power is used to defeat gravity at a given gradient.
Now THAT's clear. Ok.
I thought you were looking to somewhat theoretical gravity-only scenarios, but since you're talking about percentages you take into account drag and rolling too.
DirtyWorks said:
I stumbled upon this very nice interactive model:
http://www.gribble.org/cycling/power_v_speed.html
I can change gradient and the portion of a given speed that is dedicated to defeating gravity is nicely represented.
What it does graph nicely is all power for most cyclists goes to defeating gravity at a 6% incline. The model agrees with my limited calculations of my own performance on nearly flat course well.
BUT, it seems to me, a cyclist would need to generate far more than 300w riding 20 Kmh on a 6% grade. Maybe I'm wrong because I've never followed quantizing cyclists with great interest.
Right now, I'm only seeking a loose mathematical model that's sort of accurate.
Don't have a graph, but it shouldn't be too hard to make one (when I have some time)
The mathematical model is fairly easy.
the rider has to overcome three forces:
- gravity, which is expressed as
Fg=m*g*cos(angle) where m is the mass of everything (rider+bike), g is the free fall acceleration constant and the angle is... the angle of the road;
- rolling friction, which is expressed as
Fr=m*g*r*sin(angle) where r is the rolling friction coefficient(depends on materials);
- drag resistance, which is expressed as
Fd= d*v^2 where v is the speed and d is the drag coefficient (depends on the shape of the exposed surface)
Hence, considering that the power can be calculated as P=F*v, where F is the total force applied and v is the very same speed of the rider, we can get this object:
P=m*g*cos(angle)*v + m*g*r*sin(angle)*v + d*v^3
as you may notice, the percentage of the power spent to overcome gravity over the total is:
Pg/P = m*g*cos(angle)*v / (m*g*cos(angle)*v + m*g*r*sin(angle)*v + d*v^3) = m*g*cos(angle) / (m*g*cos(angle) + m*g*r*sin(angle) + d*v^2)
That means, it still depends from v.
You can try to play with the values and see what you get, or else you could, as mentioned earlier, use actual speeds of actual riders on each gradient.
EDIT: the site you linked looks good. I have no reason to believe there's anything wrong in that, but I didn't look carefully :V