DIY dyno calculation with an Arduino sensor circuit

[hksaaa]

TL;DR Summary

DIY dyno calculation

I would like to know if this formulae i have are good enough to make calculations.
My plan is to count time of each roll with Arduino.I don't have roller yet so i don't know moment of inertia.angular acceleration = radians/secounds
TORQUE = moment of inertia (kg∙m2) * angular acceleration
RPM = 60/secounds
HP = TORQUE x RPM ÷ 5252

 

[BvU]

Hello hks,

angular acceleration = radians/secounds

wrong dimension. [angular acceleration] = angle/time²
(analogous to [acceleration] = length/time² )

TORQUE = moment of inertia (kg∙m²) * angular acceleration

right

RPM = 60/second

No. Revolutions Per Minute, so: 1 RPM = 1/60 revolutions/s = 2$\pi$/60 radians/s

HP = TORQUE x RPM ÷ 5252

Wrong dimension [power] = energy/time

One out of four -- back to an introductory textbook or so for study

 

[hksaaa]

Hello hks,
wrong dimension. [angular acceleration] = angle/time²
(analogous to [acceleration] = length/time² )
right
No. Revolutions Per Minute, so: 1 RPM = 1/60 revolutions/s = 2$\pi$/60 radians/s
Wrong dimension [power] = energy/time

One out of four -- back to an introductory textbook or so for study

Thank you for helping me! Sadly i am not engineer so sorry.

Can you please explain me what is wrong with -> HP(horse power) = TORQUE x RPM ÷ 5252 and how it should look with my parameters?I found this on google and website looks legit.

 

[Baluncore]

Can you please explain me what is wrong with -> HP(horse power) = TORQUE x RPM ÷ 5252

You have not specified the units used for torque in the equation.
Power = torque * angular velocity;
In SI units; watt = Newton·metre * radian/second.

1 HP = 745.7 watt.
1 kg force = 9.80665 Newton.
1 foot = 0.3048 metre.
1 pound = 0.4536 kg.

 

[BvU]

HP = TORQUE x RPM ÷ 5252

I goofed writing 'wrong dimension' -- Baluncore is too kind avoiding to mention that. His post corrects me. What I should have written is a question about the context of the expression: how do you intend to use it ?

PS and I was too lazy to check on the numerics...

 

[hksaaa]

I goofed writing 'wrong dimension' -- Baluncore is too kind avoiding to mention that. His post corrects me. What I should have written is a question about the context of the expression: how do you intend to use it ?

PS and I was too lazy to check on the numerics...

In general my plan is to get time of each full roll and process/calculate data to get Horse Power and show it
in graph with given formulae.

If i got it, i should use this formula -> Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488 (and convert it in HP) with :

1. Angular acceleration(rad/s(square)) = radians/secounds(square)
2. TORQUE(N.m) = moment of inertia (kg∙m2) * angular acceleration(rad/s(square))
3. RPM = radians(per secound) * 30 / pi
4. Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488

 

[BvU]

In general my plan is to get time of each full roll and process/calculate data to get Horse Power and show it
in graph with given formulae

Good. So you get something like a list of times after startup: $t_0, t_1, t_2, t_3, ...$ for start, 1, 2, 3 ... revolutions.

Useful formulas on page 5 of the pdf in the link I gave: with $\omega_0=0$, so: from standstill, you have$$ \begin{align*}\omega &= \alpha t \\ \theta &= {1\over 2}\alpha t^2\end{align*}$$
For your observations ( with $\theta_n = 2\pi n$ ) that means $$ 2\pi n = {1\over 2}\alpha t_n^2 $$ which can be re-ordered to $$ t_n^2 = {4\pi \over \alpha }\, n $$so a plot with $n$ horizontal and $t_n^2$ should give a straight line --- in theory, and provided the torque is constant.

Experimentally it may be lesss noisy ( the analysis above heavily leans on $t_0=0$ ) to, instead, make a plot of $\ \displaystyle {1\over t_n - t_{n-1}}\$ as a function of $\displaystyle{t_n+t_{n-1}}\over 2$.

This is based on the first expression: $\ \omega = \alpha t\$ where $\omega\approx\omega_{\scriptstyle\text { average} } = \displaystyle {2\pi\over t_n - t_{n-1}} \$ for the average time $\displaystyle{t_n+t_{n-1}}\over 2$.

 

[hksaaa]

Good. So you get something like a list of times after startup: $t_0, t_1, t_2, t_3, ...$ for start, 1, 2, 3 ... revolutions.

Useful formulas on page 5 of the pdf in the link I gave: with $\omega_0=0$, so: from standstill, you have$$ \begin{align*}\omega &= \alpha t \\ \theta &= {1\over 2}\alpha t^2\end{align*}$$
For your observations ( with $\theta_n = 2\pi n$ ) that means $$ 2\pi n = {1\over 2}\alpha t_n^2 $$ which can be re-ordered to $$ t_n^2 = {4\pi \over \alpha }\, n $$so a plot with $n$ horizontal and $t_n^2$ should give a straight line --- in theory, and provided the torque is constant.

Experimentally it may be lesss noisy ( the analysis above heavily leans on $t_0=0$ ) to, instead, make a plot of $\ \displaystyle {1\over t_n - t_{n-1}}\$ as a function of $\displaystyle{t_n+t_{n-1}}\over 2$.

This is based on the first expression: $\ \omega = \alpha t\$ where $\omega\approx\omega_{\scriptstyle\text { average} } = \displaystyle {2\pi\over t_n - t_{n-1}} \$ for the average time $\displaystyle{t_n+t_{n-1}}\over 2$.

Thank you for well explained!

My plan was to get time for each full roll, calculate HP for given time and put in curved graph like below(its with fake data just to test if its working) and repeat for let's say 15 secounds to get full graph.

Are my formulae correct for this situation?

1. Angular acceleration(rad/s(square)) = radians/secounds(square)
2. TORQUE(N.m) = moment of inertia (kg∙m2) * angular acceleration(rad/s(square))
3. RPM = radians(per secound) * 30 / pi
4. Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488

 

[BvU]

I think so. Is 'time for each full roll' the same thing as 'time for a full revolution' ?

Found your funny 5252 coefficient in the engineering toolbox -- glad I move in an SI units world .

 

[hksaaa]

I think so. Is 'time for each full roll' the same thing as 'time for a full revolution' ?

Found your funny 5252 coefficient in the engineering toolbox -- glad I move in an SI units world .

Yes its same thing!

It was first result in my google search.

Thank you once again, appreciate your help!
Will post update when i try it(missing sensor at the moment).