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I have a power curve relating to a turbo trainer I use on my bike.

I save my sessions on a website called strava using speed, cadence, heart rate and time.

Using the power curve I've been given I wish to calculate the gradient of the curve, providing me with a power calculation for a given speed.

$\qquad\qquad\qquad$enter image description here

I've got a rough table of figures that can be used to plot a graph, but I can't remember the maths I need to work out the formula.

$$ \begin{array}{c|r} \text{speed} & \text{power} \\ \hline 5 & 25 \\ 9 & 50 \\ 12.5 & 100 \\ 17 & 200 \\ 20.5 & 300 \\ 23 & 400 \\ 25 & 500 \\ 27 & 600 \\ \end{array} $$

Thanks for any advice you can give me.. even better if you can provide the completed formula.

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2 Answers 2

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The power curve seems to be approximately of the form $P=cv^{2}$, where $P$ is the power, $v$ is the speed and $c$ is a constant that we need to find. To reduce the error I computed the following mean based on your figures $$\begin{equation*} c=\sqrt[8]{\frac{25}{5^{2}}\frac{50}{9^{2}}\frac{100}{12.5^{2}}\frac{200}{ 17^{2}}\frac{300}{20.5^{2}}\frac{400}{23^{2}}\frac{500}{25^{2}}\frac{600}{ 27^{2}}}\approx 0.75. \end{equation*}$$

So the approximate equation of the form $P=cv^{2}$ is $$\begin{equation*} P\approx 0.75v^{2}\qquad \text{(}P\text{ in Watt and }v\text{ in mph),} \end{equation*}$$

or $$\begin{equation*} v\approx 1.16\sqrt{P}. \end{equation*}$$

This approximation gives

$$\begin{eqnarray*} v(600) &\approx &28.4 \\ v(300) &\approx &20.1 \\ v(100) &\approx &11.6. \end{eqnarray*}$$

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excellent. I'd tried plotting against p=cv^2 and couldn't get beyond that in terms of a relationship. Thank you. –  Simon Forster Sep 3 '12 at 15:31
    
@SimonForster You are welcome. –  Américo Tavares Sep 3 '12 at 15:33
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"Using the power curve I've been given I wish to calculate the gradient of the curve, providing me with a power calculation for a given speed."

If I understand right, the curve itself already gives you the power as a function of speed -- you don't need to do anything with the gradient. And I think that the curve itself was derived empirically by the company so there is not necessarily a formula to "work out" other than just interpolating between the points.

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well I assume as there is a graph with a curve on it there is an equation to work out a point on this line using a given value of x or y. what I want to do is use this equation to calculate on the fly the power at any given speed.. so that I can include this with workout information –  Simon Forster Sep 3 '12 at 14:11
    
just because someone printed a curve on some paper or a website doesn't mean there's a nice corresponding equation; maybe the jetfluid pro resistance equation would involve complicated fluid dynamics calculations if you wanted to get it using math instead of using empirical measurements. but you could approximate the curve using a piece of a parabola or something. –  binn Sep 3 '12 at 15:22
    
yes it does.... –  Simon Forster Sep 3 '12 at 15:30
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