# How to determine the angular velocity of the platform as a function of motor rotations per second?

Hello all of you engineering geniuses. Awhile ago a very talented individual on here helped me work out the conceptual mechanics of this design. As you can see from the model, there is a lead screw assembly that pushes a rod forward which elevates the platform. The total travel is 0° to 60° (theta). The equation shown is the linear travel of "Xblock" as a function of theta. If you were to plot this function, you will see it is not quite linear. The terms, Wp, Ht, Hp, and Lrod are all constants. I am trying to determine an equation for angular velocity of the platform, which is actually decelerating as Xblock moves forward at a constant velocity. Of course the linear travel is easy enough, simply the rps of the motor x lead x time. The lead is 8mm but the lead and rps can be treated as constants. I could simply take the derivative of function, but it's a function of position as a function of another position, not time, so I'm not entirely sure where to go from here. • looks like you can derive dx/dϑ directly from your equation (e.g. with Mathematica, or numerically if it makes your life easier) invert it to get dϑ/dx, multiply that by dx/dt to get dϑ/dt Feb 14, 2021 at 18:04

As PeteW mentioned you can differentiate both sides of the equation with respect to t, and then solve for $$\dot\theta(t)$$.

If I done correct the calculations, you'll end up with

$$\dot\theta(t) = -\frac{X'(t)}{a(\theta(t)) + b(\theta(t))}$$

where:

• $$a(\theta(t))= -H_{p} \cos(\theta(t))+ \frac{1}{2} W_{P} \sin(\theta(t))-L_{rod} \arcsin\left(\frac{H_T-H_{p} \cos(\theta(t))+\frac{1}{2} W_{P} \sin(\theta(t))}{L_{rod}}\right) \sin(\theta(t))$$
• $$b(\theta(t))= \frac{\cos(\theta(t))}{\sqrt{1-\frac{\left(H_T-H_{p} \cos(\theta(t))+ \frac{1}{2} W_{P} \sin(\theta(t))\right)^2}{L_{rod}^2}}} \left(\frac{W_{P}}{2} \cos(\theta(t))+H_{p} \sin(\theta(t))\right)$$
• $$X'(t)=\frac{2\pi n}{60} l_{thread}$$: the velocity of the cart:
• $$n$$ is the revolutions per minute [rpm]
• $$l_{thread}=8[mm]$$ is the thtread size in [meters]

After, all the simplifications the angular velocity can be calculated as a function of $$n [rpm]$$, and $$\theta [rad]$$

$$\dot\theta(n, \theta) = -\frac{2\pi l_{thread}}{60(a(\theta) + b(\theta))}n$$

where:

• $$a(\theta)= -H_{p} \cos(\theta)+ \frac{1}{2} W_{P} \sin(\theta)-L_{rod} \arcsin\left(\frac{H_T-H_{p} \cos(\theta)+\frac{1}{2} W_{P} \sin(\theta)}{L_{rod}}\right) \sin(\theta)$$

• $$b(\theta)= \frac{\cos(\theta)}{\sqrt{1-\frac{\left(H_T-H_{p} \cos(\theta)+ \frac{1}{2} W_{P} \sin(\theta(t))\right)^2}{L_{rod}^2}}} \left(\frac{W_{P}}{2}\cos(\theta)+H_{p} \sin(\theta)\right)$$

• Well look who it is! Seems like you are always the man for the job. You should be pleased to know that I used your original equation to derive a 3rd order polynomial (based on hard values for Hp, Wp, etc.) and am utilizing that in LabVIEW to achieve positioning. I'm getting great accuracy, precision, and resolution even with this model made of parts thrown together. I could have used that polynomial for these calculations but it's still a best-fit line, so some data points are off, albeit only by a few negilible milliseconds. Once again, amazing work. Thank you. Feb 15, 2021 at 15:23
• Hi, First of all thanks for the kind words. Its my turn to say that your build is very impressive, and you have it fully instrumented. I am quite interested in how you control the Labiew.. do you use position control or do you control velocity? In any case, I wanted to remind you that, as you increase the velocity you will see greater differences, because the dynamic effects were not considered in the derivation.
– NMech
Feb 15, 2021 at 18:56
• imgur.com/VQ5nwrX Feb 16, 2021 at 18:47
• imgur.com/LZ8P5RK Feb 16, 2021 at 18:48
• imgur.com/m8NPapC Feb 16, 2021 at 18:48