UKC

Does riding up a hill of a given gradient in a given gear (road bike) equal riding up a steeper hill in a lower gear, all things being equal?

I know there are formulae for such things but they look massively complicated for someone with an interest in maths but no real talent.

I'm interested in the science behind it rather more than anecdotal. Instinct says that power is power and should be transferable but some figures would interest my inner geek.

In reply to Stuart (aka brt):
The crude science says that the power used will be the same so long as the same weight is lifted the same height while travelling the same distance.

In reality it's so much more complicated because efficiency changes at different power outputs and gear ratios.

In reply to gethin_allen:

I'm not sure that's 100% correct.

If we ignore the fact that the bike and it's rider have different efficiencies at different power outputs it's actual quite simple.

Let's imagine a hill with one steep side and one shallow side. Both sides gain the same amount of height. The energy required to get to the top of the hill by either route is exactly the same, it is the mass of the cyclist and bike times by the height gained times by a constant.

In terms of power (which is how much energy you're putting out per second) it all depends on how fast each route is done. If I ride up the shallow side in a high gear I'll be travelling faster but gaining less height with every meter of road. If I ride up the steep side in a smaller gear I'll be travelling slower but gaining more height with each meter of road and thus the power is the same.

For a power output of x both routes take the same amount of time and feel just as difficult.

In reality it's much more complicated than that but that's what I reckon the physics says.

In reply to Greasy Prusiks:

Maybe the " same distance" thing is wrong/confused, I was trying to factor in wind/tyre rolling resistance but failed. Because obviously you can't climb the same height at a lesser gradient traveling the same distance. There's a gcn how to calibrate your power meter vid on YouTube that's quite interesting regarding this subject

In reply to gethin_allen:

From my meagre understanding of science, the inconvenience of variables seem to get left out. Which in this case, seems sensible (though of not much real world value).

Just playing with the concept in my head really. Wish I had the ability to crunch the numbers. As it is I just tend to throw myself up the hills.

I presume it's possible, all things being equal (or ignored), to work out a gradient at which a rider with a given power threshold over a given time and a given gear ratio, would grind to a halt?

In reply to Stuart (aka brt):

I think you could work out how far up a hill a rider would get if they can sustain x watts for y minutes. It would be independent of the steepness of the hill.

In reply to gethin_allen:

I definitely see what you're getting at. I wasn't quite sure about the power used bit that's all.

In reply to Greasy Prusiks:


I assume the maths is fairly straightforward?

In reply to Stuart (aka brt):

Depends if the hill is on a treadmill

In reply to Stuart (aka brt):

Yeah I could show you how to estimate it if you have numbers for you're power output?

In reply to Greasy Prusiks:
I don't. It was more conjecture on how steep would too steep. Googling around suggests that all things considered 60% gradients are the human/bike (friction) limit.


But thanks anyway.

In reply to Stuart (aka brt):

OK. That sounds a steep climb! I reckon you'd go A over T at some point.

In reply to Greasy Prusiks:


Haha. Found something Keith Bontrager had worked on. It did require a special frame to set the centre of gravity just so.

In reply to Stuart (aka brt):

Does what equal what? Power output at a given pedal rate? Overall efficiency of food > work? The cycling experience? The effort of maintaining a constant vertical climb rate? The torque required to do a wheelie? Something else?

Steeper hill at constant power means slower speed as more work goes into going up, meaning less goes in to going forwards. Unavoidable losses are friction and air resistance relating to going forwards in a more than linear way, so going slower reduces the total energy lost to these. Although weight distribution changes may alter this.