# How to calculate time from different instantaneous velocities and accelerations?

I have a closing door, that moves from angle 75 to angle 5. Torque acting on that door at each angle is known -of course differs at each angle- I want to calculate the time required for that door to close, knowing that it starts with angular velocity $\omega(t=0) = 0$ and angular acceleration $\alpha(t=0) = 0$ at angle 75°.

I tried to calculate angular acceleration at each angle as: $$\alpha = \tau / I,$$ where $I$ is the moment of inertia of the door.

Acceleration is different at each angle. From instantaneous acceleration and knowing the distance between each calculation positions ($\Delta angle = (\pi/180)$) I could get angular speed at each position as: $$\omega_\text{final}^2 = \omega_\text{initial}^2+2\,\Delta (\text{angle})\,\alpha_\text{average}$$

Then I calculate ($\Delta time$) for each as: $$(\Delta t) = (\omega_\text{final}-\omega_\text{initial})/\alpha_\text{average}.$$

Then I get the summation of all ($\Delta t$) to get the total time required for the door to close from 75 to 5 degrees but it was completely different from the actual measured time :(.

Is there any thing wrong about that calculation procedure? What is the right way to calculate time for such problems?

• If the applied force was different than what was calculated, it could cause the difference. And how different is "totally different" that you measured? Jan 15, 2015 at 13:25
• no , it is the same applied force , cuz i'm validating my results and comparing them with the valid model , time was 0.68 in the valid model , but in my model it was 0.079 seconds !! Jan 15, 2015 at 14:16
• Based on your starting conditions, $\alpha$ must become non-zero at $\ang{75}$ at some time, t>0. Do you know what that time is? Otherwise, the door never moves. Jan 15, 2015 at 21:21

If you know the torque and hence the acceleration as a function of position then

$$\frac{{\rm d}^2\theta}{{\rm d}t^2} =\frac{{\rm d}\omega}{{\rm d}t} = I^{-1} \tau(\theta)$$

$$=\frac{{\rm d}\omega}{{\rm d}\theta} \omega = I^{-1} \tau(\theta)$$

$$\int \omega \,{\rm d} \omega = \int I^{-1} \tau(\theta) \, {\rm d} \theta$$

$$\frac{1}{2} \omega^2 +C = \int I^{-1} \tau(\theta) \, {\rm d} \theta$$

Now you have the calculated speed as a function of angle given some initial conditions and a numerical integral

$$\omega(\theta) =\sqrt{2 \left( \ldots \right) }$$

A second integral is needed to get to the time needed to reach a certain speed.

$$t = \int \frac{1}{\omega(\theta)}\,{\rm d} \theta$$

## Edit

So if at one instant you measure/calculate the rotational speed and acceleration as $\omega_1$ and $\alpha_1$ then you can estimate the speed at the next step $\omega_2 = \omega_1 + \Delta \omega$ if you know the acceleration change $\alpha_2 = \alpha_1 + \Delta \alpha$. By assuming linear acceleration curve between the points you arrive at

$$\frac{1}{2} \omega^2 - \frac{1}{2} \omega_1^2 = \int \limits_{\theta_1}^\theta \alpha(\theta)\,{\rm d}\theta$$ which simplifies to

$$\frac{1}{2} \omega^2 - \frac{1}{2} \omega_1^2 = \alpha_1 (\theta-\theta_1) + \frac{\Delta \alpha}{2 \Delta \theta} (\theta-\theta_1)^2$$

$$\frac{\Delta \omega}{2} (2 \omega_1+\Delta \omega) = \alpha_1 \Delta \theta + \frac{\Delta \alpha}{2} \Delta \theta$$

$$\boxed{ \Delta \omega = \sqrt{\omega_1^2+\Delta \theta (2 \alpha_1 + \Delta \alpha)} -\omega_1 }$$

This can also be expressed as

$$\omega(\theta) = \sqrt{ \omega_1^2 + 2 \alpha_1 (\theta-\theta_1) + \frac{(\theta-\theta_1)^2 \Delta \alpha}{\Delta \theta}}$$

The time for the step $\omega_1 \rightarrow \omega_1 + \Delta \omega$ is estimated by

$$\Delta t = \int \limits_{\theta_1}^{\theta} \frac{1}{\omega(\theta)}\,{\rm d}\theta$$

$$\Delta t = \sqrt{\frac{\Delta \theta}{\Delta \alpha}} \ln \left(1+ \frac{\sqrt{\Delta \alpha}(\Delta \omega+\sqrt{\Delta \alpha \Delta \theta}}{\alpha_1 \sqrt{\Delta \theta}+\omega_1 \sqrt{\Delta \alpha}} \right)$$

The above is simplified a bit if you consider the time constant $T = \sqrt{\frac{\Delta \theta}{\Delta \alpha}}$. When $\Delta \alpha \neq 0$ and $T \neq \infty$ then $$\boxed{ \Delta t = T \ln \left(1+ \frac{\Delta \theta + T \Delta \omega}{T (\omega_1 + T \alpha_1)} \right) }$$

Otherwise when $\Delta \alpha=0$ and $T=\infty$ $$\boxed{\Delta t = \frac{\Delta \omega}{\alpha_1}}$$

• unfortunately , torque is not a function of position , it is measured at each position individually :( Jan 15, 2015 at 18:12
• As long as you have torque data as function of position then the above applies. Just replace the integrals with discrete sums. Jan 15, 2015 at 20:25
• so , this procedure is not valid :( Jan 17, 2015 at 14:44
• I think it is. Torque is measured at each position and thus is treated like function. Jan 17, 2015 at 22:19
• See edits above for more details on how to do this with a time series. Jan 17, 2015 at 23:05