Friday Math Quiz

Ok everyone, I should be able to do this, but it is eluding me today. I have been asked to change a servo axis such that it has a much slower accel and decel rate to save wear and tear on other hardware. Currently the rates are at 100% of max, so I'm assuming infinity. I need to alter the accel/decel rates and they will be set at 5 in/s^2 on both ends. What I need to do is determine the appropriate new max speed for the axis to eliminate the loss of cycle time due to the slower accel/decel rates.

So, assuming that before the axis traveled at 4.5 in/sec for 21.5 inches, I can get the cycle time. I need to set the new velocity max such that I accel to this new velocity, remain at it, and decel back to zero in the same amount of time. For some reason, I can not set this up as an inequality and get the minimum speed to accomplish this without screwing something up. Anyone want to take a quiz today?

Russ

I'll opt for multi choice answer C

Ha Ha,

In all seriousness, this is something I should be able to represent mathematically and calculate. I have set the max speed already, using the dummy method of guess a value and check, then guess higher if the total cycle time is larger than the original, but I am ashamed to say I had to use that method. I was just hoping someone here could straighten me out on a Friday.

I came up with:

(215 - Sqrt(11395))/18 = Approx 6.014 in/sec

Let:
a = length of time for accel/decel
t = time running at constant velocity
x = constant velocity to be found

Total distance travelled given as 21.5 (=43/2)

Area under velocity time graph = 43/2

Thus ax/2 + tx + ax/2 = 43/2;

Re-arrange gives (a+t)x=43/2

Accel rate given as 5, so x=5a, sub in above gives

5a(a+t)=43/2

Total cycle time must remain constant, so

Original cycle time 21.5/4.5 = 43/9 must equal a+a+t

Thus 2a+t=43/9 -> t=43/9 - 2a, sub in above

Gives 5a(a + 43/9 -2a)=43/2

Simplify and expand out:

-5aa + 215/9a - 43/2 = 0

Solve the quadratic in a and then x=5a

ow!

I tried once and realized my approach wouldn't work. I need a bit more coffee before I attempt again.

Thanks for the headache :*)

Russ,

The final velocity v of an object which starts with velocity u and then accelerates at constant acceleration a for a period of time (Δt) is:



In your case we know the initial velocity u = 0, and a (acceleration) = 5 in/sec^2, but we don't delta-t, the time it takes to accelerate (and decelerate) the mass. That is dependent on the mass and the force applied.

We do know that the total travel time is 21.5/4.5 or 4.778 seconds, so we can make an assumption that the deceleration time will be at least equal to the acceleration time (it may be different), and that we accelerate 1/2 of the travel time, then decelerate the remaining 1/2. Now we make a logical assumption of 2.389 seconds acceleration and 2.389 seconds deceleration. Then we can compute the final maximum POSSIBLE velocity:

V = 0 in/sec + (5 in/sec^2) X 2.389 seconds = 11.942 in/second

So the device COULD start at 0 speed at 0 time, accelerate at 5 in/sec^2 for 2.389 seconds, reaching a velocity of ll.942 inches per second, then decelerate for 2.389 seconds back to 0 speed, with total time 4.778 seconds and distance of 21.5 inches.

Thanks L D[AR2,P#0.0]

L D[AR2,P#0.0]'s answer is correct. This is what I was attempting to do, integrate under the velocity time curve. The second variable for some reason was just blowing me away though.

Lancie, your answer is only useful if there is no period of constant velocity. Essentially the move profile was altered from a flat constant velocity to a period of accel to speed, constant speed, and decel back to zero. I actually set the max speed to 6.25, as this reduced the cycle time slightly, but I wasn't able to just do this based on a calculation like I should have been. My goal was to set this such that the cycle time would be basically unaffected due to the accel/decel rate changes. While your approach would work, I'd also have an axis moving at almost 3 times the speed it used to, which is not desireable. Beyond that, a quick check reveals your answer to be incorrect. The formula for constant acceleration and change in position is delta X = ViT + .5aT^2. Since I'm starting at zero, all that matters is the .5aT^2. Using a=5 and T=2.389 seconds, then multiplying by two for the accel and decel sum, I get a travel of over 28 inches.

Thanks all, this one stumped me completely for a bit.

Thanks again.

Russ

Russ, what can I say? I don't see that you mentioned that it had to run at a constant velocity for some time period. Nor did you mention that the speed was constrained, only that the acceleration was constrained to 5.

My answers can only be as good as your description of the problem....

See my edits above

Lancie,

See my edits above. Something didn't look correct with your answer, and it works out to have excess travel. While it would have worked in the context of time, the drive would never ever reach it's at speed state if the move were only over 21.5 inches. Thanks again to all.

Russ

I came up with the same thing as L D[AR2,P#0.0] but on a slightly different path.

My thoght was:

2*d_a + d_cv = 21.5

where:
d_a is accel distance
d_cv is constant speed distance

d_a is simply V²/(2*a)
where:
V is constant speed velocity
a is accel rate

d_cv is V*t_cv

We know the total original move time was 21.5/4.5 or 4.77777 seconds.

t_cv = T_tot - 2*t_a

t_a is V/a

So, in total, we have:
(Equ A)
(2*(V²/(2*a) ) + (V * (4.77777 - (2*V/a))) = 21.5

or

V²/5 + 4.777777*V - (2 * V² / 5) = 21.5
4.77777 * V - V² / 5 = 21.5
0 = V² / 5 - 4.777777*V + 21.5

Now use the quadratic formula on this equation to get the roots. You will get two; 6.014074 and 17.8746.

If you substitute these into the Equ A above you will find that only one of these gives you the right result. 17.8746 will produce a negative value for constant speed time, which is bad.

OK, that was kind of fun.

Keith

You guys are close

Actually, I have these kind of problems already solved. I just need to find the right Mathcad work sheet and plug in the numbers.
Russ, the trial and error method works. At least it provides a sanity check.
There was a thread started by Norm Dzeidic about a simliar topic where the time to accel, time at constant velocity and the time to decel are all the same. This yield similar results as in Russ's problem. I worked them out once and have these memorized so I can get a quick idea of what the velocities and accelerations need to be.

v=1.5*d/t
a=4.5*d/t^2
where t is the total time.

After I work this out using radcon math ( nuke speak for quick estimate in my head ) I then pull out the Mathcad and see how close my estimate was.

Interesting

Peter,

It's pretty cool that you can memorize those things. I never bothered to, although today appeared as an example of why I should. I always have too damn much fun trying to figure it out again though.

One question, does anyone ever specify their move profile to be broken out into thirds as far as time or position is concerned? It would seem to me that the load #'s and permissible torque and or force values that were deemed acceptable would be the starting point. At least, that was the case here.

This whole issue was brought up by a linear actuator which had a bunch of slop in it from 2 years of service. When I mentioned that I believed we could ease the strain on the actuator by adjusting these values without affecting cycle time, management thought I was a hero.

I'm new to system design, as I only have a couple years experience. My extensive experience has been in maintenance of these systems and troubleshooting. In my limited experience, I would think that starting with the hardware limitations would be the foundation of creating the "optimum" move profile. So, to restate my question, is it fairly common in anyone else's experience to have the principal focus put on obtaining certain time or position percentages for accel/decel? Just a curiosity question more than anything.

Peter - Buy saying we are close implies there is a mistake or the like. What exactly are you trying to say ?

I was just kidding about the precision of the calculation, that's all

One question, does anyone ever specify their move profile to be broken out into thirds as far as time or position is concerned? It would seem to me that the load #'s and permissible torque and or force values that were deemed acceptable would be the starting point. At least, that was the case here.

Click to expand...

No, not really.

First you must realize that this is what I do when I am not goofing off here. If it was your job to come up with motion control algorithms then you would definitely have the simple ones memorized.

The reason why I memorize this is that I often get calls from customers that need to move from point A to point B is a certain amount of time just as in your question. The 1/3 1/3 1/3 formula lets me make a quick estimate of what the requires will be in a few seconds on the phone. It is a good place to start. Obviously there are many combinations of the velocity,acceleration and time that will meet your needs. The most efficient motion profile is a parabola however the parabola does not provide for smooth starts and stops. A 1/3 1/3 1/3 profile is not quite as efficient but one can convert the linear ramps to s curve ramps and have a very smooth motion profile that is relatively efficient.

I'm no motion control expert, but all anyone here used was high school physics kinematics (derived from very basic calculus). How are there "many combinations of the velocity,acceleration and time that will meet your needs"? As I see it, there is one meaningful solution assuming constant acceleration. Sure there are more combinations if you have a non-zero jerk rate (change of acceleration) and subsequent time derivatives, but you could show greater acceleration rates at some with mean value theorem, which is exactly what OP is trying to minimize.

Am I missing something? This is strictly based on minimizing the maximum acceleration rate. I don't understand the parabola as the "most efficient" motion profile, but could see that a controlled jerk at the beginning at end of a smooth transition as easier on your equipment - the real goal of the post.

Peter Nachtwey said:

Obviously there are many combinations of the velocity,acceleration and time that will meet your needs. The most efficient motion profile is a parabola however the parabola does not provide for smooth starts and stops. A 1/3 1/3 1/3 profile is not quite as efficient but one can convert the linear ramps to s curve ramps and have a very smooth motion profile that is relatively efficient.

Click to expand...