# A maths question...

G-Wiz
Posts:

**261**
If it takes 10 minutes to get up Box Hill in a 34-26 gearing, at an average cadence of 85 rpm

A) What rear cog size would it need to get up in sub 8 minutes at the same cadence?

B)Assuming a 100kg rider on an 8kg bike, what increase in power does the above require?

Show your workings :-)

A) What rear cog size would it need to get up in sub 8 minutes at the same cadence?

B)Assuming a 100kg rider on an 8kg bike, what increase in power does the above require?

Show your workings :-)

0

## Posts

1,282A) 10/8 * 26 = 32.5 tooth sprocket, ask at halfords for one

Too complicated, increase in power output isnt directly proportional to the increase in speed, loads of other factors like air resistance isnt directly proportional to speed, efficiency of body also not proportional.

Would be a lot of power though for your theoretical 100kg fat b******

10,8621.308 * 10/8 = 1.635

34/1.635 = 20.78 = rear sprocket size.

Twitter

Flickr

780http://cozybeehive.blogspot.com/2009/07 ... ained.html

112You might want to check that... you're saying that a *larger* rear cog would get you up the hill quicker at the same cadence!

1,282My mistake :oops: .....it was late evening.....8/10 * 26 = 20.8

6,1322) I don't know Box Hill and how steep it is - if it's > 8 - 10%, speeds would be low enough that the significance of wind resistance drops off. In that case you can use soemthing like Allen Lims formula:

http://www.trainingbible.com/joesblog/2 ... rmula.html

"Dr. Lim's formula to estimate the power necessary to climb a hill:

bike + rider weight (kg) x 9.8 x elevation gain (meters) / time (seconds) = power (watts). Add 10% for rolling and air resistance."

As you're only interested in the change in power to achieve a quicker time, the increase in power is simply proportional to the reduction in tiime

Power (8mins) = Power (10 mins) * 10/8 , so 25% more power

However, the flatter the hill is, the greater the imapct of wind resistance and ultimately it becomes the dominating component. In that case the power will go up as the square of the speed increase so you'll need ~56% more power.

The "real" answer is somewhere between the two.

261Nothing theoretical about the fat censored , although 'he' is now down to 98Kg already, but that would make the maths even more daft.

That equates to speeds between 13-15kmh so not a huge impact of air resistance.

I'm hoping to get down to low 90kg's over the season, already only 1Kg off last season's best when I was almost under 9 mins. That was the first year I did any competitive riding, so I'm expecting to cane those results now I've a bit more structure and some targets in place.

Thanks especially for Dr Lim's formula, I reckon that's probably saved me needing a powermeter for a little while.