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Trapezoidal Motion Profile

I'm working on equations used to solve non standard motion profiles with limited known values. In this example, I'm having trouble solving equations where acceleration and deceleration is different. It's really just calculating trapezoidal value unknowns.
Given:
Total Distance (Area)
Total Time (Base Length)
Acceleration (Slope of Accel Triangle)
Deceleration (Slope of Decel Triangle)
Calculate:
Vmax (Height)
T1 - Accel Time
T2 - Constant Velocity Time (Top Length)
T3 - Decel Time


I'm working on problems that require different acceleration and deceleration rates and I would like to understand the mathematics behind solving the issue with only known values. I have a software that will calculate the outputs but don't know quite how the math works.
For Example (Known Values):
Total Distance = 1000 units
Total Time = 4 sec
Accel = 1000 units/sec2
Decel = 500 units/sec2
Output
Vmax - 279 units/sec
T1(accel) - 0.27924 sec
T2(constant) - 3.16228 sec
T3 (decel) - 0.55848


Can someone help with the formulas need to calculate these unknown values.

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From the area of the trapezium you have $$(t_3+4)\frac V2=1000$$

From the suvat formula $v=u+at$ you also have $$V=1000t_1=500t_2$$

And, of course, $$t_1+t_2+t_3=4\implies t_3=4-\frac{3V}{1000}$$

This leads to the quadratic equation $$\left(8-\frac{3V}{1000}\right)\frac V2=1000$$

Solve this and you get $$V=279.24$$ as well as another, spurious, solution. The other results then follow.

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  • $\begingroup$ To make sure I follow you correctly. Trapezoid Area = (Base + Top)*Height/2.... Thus don't you mean to use t2 vs t3 since it's the top constant velocity section? I'm pretty sure I see the rest and it was the quadratic part that was throwing me off... $\endgroup$ – Zoul007 Sep 2 '16 at 2:20
  • $\begingroup$ In your diagram $t_3$ is the length of the top $\endgroup$ – David Quinn Sep 2 '16 at 8:14
  • $\begingroup$ You are correct (T3) is the constant speed section, I forgot which diagram I uploaded. ;( $\endgroup$ – Zoul007 Sep 7 '16 at 13:40

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