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Submitted by aaronkeit on Thu, 10/27/2011 - 13:30

A while ago I asked for help calculating the speed required for a trapezoidal move if the duration desired and the acceleration/deceleration were known. Member RTopholm (Richard) helped me and gave the following formula, assuming accel and decel are not the same:

v = 0.5*((a*c)/(a+c)) * (t - sqrt(t^2 - 2*s*(1/a+1/c)))

I'm just getting around to implementing this now, and I am trying to figure out how I will need to modify the formula if my initial speed is not zero, but the axis is already moving at the time this move begins. This move will be ending at a speed of zero, but starting when the axis reaches a specified trip point as a certain jog speed.

Any help is greatly appreciated.

Aaron

Comments 5

Galil_MarissaT on 10/27/2011 - 16:55

The equations you are looking for are the kinematic equations.

[img]http://www.physicsclassroom.com/class/1dkin/U1L6a1.gif[/img]

To solve them, assume at first that you you have three separate profiles: the acceleration portion, the slewing portion, and the deceleration portion. The kinematic equations above govern each of these individual profiles. Each of the profiles will have separate times, t (t1, t2, t3) and distances, d (d1, d2, and d3) that they travel. What will over lap is that the final velocity of the first profile will be the slewing velocity in the second, and the slewing velocity in the second will be the initial velocity in the third profile. Solving these sections using the kinematic equations you'll obtain three different equations. To unite them, we have to make an assumption such that the time of all three add up to the known t we want or the distance of all three add up to the known d we want. In fact, you'll need to make both of these assumptions to complete the equations and remove all the unknowns. One you remove all the unknowns, simply solve for the slewing velocity, as desired.

aaronkeit on 10/27/2011 - 20:08

Marissa,

Thanks for the tips, but I think I'll need a little more if possible. I understand the Kinematic equations individually, and solving for the distance seems straightforward enough, but when I try to isolate the Final velocity on one side of the equation I get stuck... it's been too long since I've done this kind of math and my skill aren't that sharp anymore.

Am I correct that, in my situation where I start at one velocity (Vi) and end at another (Vf)

d = (Vf)*t - 0.5*(Vi-Vf)^2/a - 0.5*(Vf)^2/c

From there, when I try to solve for (Vf) I get stuck at some point with a long complex equation that I am sure I am going about solving the wrong way. Any more help in getting the equation I need will be greatly appreciated :)

Thanks much,

Aaron

Galil_MarissaT on 10/28/2011 - 16:40

I have e-mailed Aaron the solution to the mathematical problem in .pdf form.

If anyone is interested in seeing the solution, please e-mail me at marissat at galilmc. dot com

maee on 10/27/2017 - 05:39

Why is the post not visible to me?

KushalP on 10/30/2017 - 13:46

It should be visible to you.