How to calculate HP knowing flywheel weight So I've seen a guy on Youtube that has an RC nitro engine. He attached a quite big flywheel of known weight to its output, and then by measuring the time that it takes to reach max rpm he was able to calculate the HP (horsepower) and torque. Now, I want to know the math behind it. I guess that it has something to do with inertia but I'm not a physics genius. I suspect that he also knows the diameter of the flywheel. So if for example we have a flywheel that is 50cm in diameter and weighs 5kg and the time it takes to rev it up to 8k rpm is 10s then how much hp would it be? If someone is willing to explain it to me I'll be happy.
 A: To my knowledge, it is not possible to calculate the engine's power from the information that you provided. Several things are missing:

*

*The answer depends on the geometry of the flywheel, more specifically it's distribution of mass. Once you know that, you can use one of these formulas to calculate it's moment of inertia $I$ for the given axis of rotation. In case none of the formulas fit because the mass distribution $\rho$ of the wheel differs from all the cases mentioned in the linked article, $I$ can be calculated by integration using the formula for a single mass point, which is how the linked formulas are obtained, too. For the following, I will assume that the moment of inertia is known.


*An engine cannot have constant torque and constant power output throughout an acceleration, because rotational energy (the time-integral of power) and angular momentum (the time-integral of torque) are not proportional to each other. So at least one of those two will be time dependent. I will show how to calculate both assuming that torque is constant:

*

*Let $f = 8 \cdot 10^3~\text{rpm}$ be the final rotational frequency of the flywheel. This corresponds to an angular velocity of
$$
\omega = 2 \pi f~.
$$
The angular momentum of the flywheel is then
$$
L = I \cdot \omega~,
$$
and it's rotational energy
$$
E = \frac 12 I \omega^2~.
$$
If the torque $M$ is constant during the acceleration and the latter took $T = 10~\text{s}$, it holds
$$
M = \frac L T = \frac{I\omega}{T}~.
$$
Then the angular velocity as a function of time $t$ is $\omega(t) = M t/I$, so
$$
E(t) = \frac 12 \frac{M^2 t^2}{I}~,
$$
and the engine's power output as a function of time becomes
$$
P(t) = \partial_t E(t) = \frac{M^2t}{I}~.
$$


*Similarly, if the power output of the engine is constant during the acceleration, the power can be obtained by dividing the final rotational energy by the acceleration time and use that to calculate $\omega(t)$, which then yields $L(t)$ and finally, by differentiation, $M(t)$.


*In case neither torque nor power are constant, there will hardly be a way to calculate them and probably they will have to be measured or looked up in a specification the engine's manufacturer might provide along with it.
