How to calculate min acceleration between two points of different velocities

**calltronics** · August 6th 2012, 12:35 PM

Hi all.

Start Point A at a velocity of 100mm/sec
Finish Point B at a velocity of 40 mm/sec

Distance to travel = 300mm
Time to do it in = 2 seconds

Now obviously we have to apply an excessive acceleration to a "Point C" (between A and B) to achieve the 300mm movement within the 2s time.

So we have an acceleration Point A to Point C
Then a deceleration from Point C to Point B

Point C will be the peak velocity to achieve the distance in the time constraints

Can anyone point me in the direction to solve the minimum value of acceleration, such that acceleration = deceleration

Cheers

**skeeter** · August 6th 2012, 01:20 PM

Start Point A at a velocity of 100mm/sec
Finish Point B at a velocity of 40 mm/sec

Distance to travel = 300mm
Time to do it in = 2 seconds

given those constraints ...

$v_0 = 100 \, mm/s$

$v_f = 40 \, mm/s$

$\Delta t = 2 \, s$

would require a constant acceleration of $a = -30 \, mm/s^2$

however, the distance traveled under that constant acceleration would be $\Delta x = 140 \, mm$ , much less than the given $300 \, mm$ of travel ... any other given information?

**ebaines** · August 6th 2012, 01:31 PM

The issue is that you have over-constrained the problem. As skeeter points out if you want the deceleration to take 2 seconds it only requires 140 mm of distance to affect the change of velocity that you want. Alternatively if you want to use the full 300 mm of distance then you could use the formula a = (vf^2 - vi^2)/2d to get a = -14 mm/s^2, but this requires time of 4.3 seconds to make the change in velocity.

**calltronics** · August 6th 2012, 01:34 PM

sorry further explanation added to question

**skeeter** · August 6th 2012, 01:50 PM

Originally Posted by calltronics

Hi all.

Start Point A at a velocity of 100mm/sec
Finish Point B at a velocity of 40 mm/sec

Distance to travel = 300mm
Time to do it in = 2 seconds

Now obviously we have to apply an excessive acceleration to a "Point C" (between A and B) to achieve the 300mm movement within the 2s time.

So we have an acceleration Point A to Point C
Then a deceleration from Point C to Point B

Point C will be the peak velocity to achieve the distance in the time constraints

"peak velocity" ? ... if you mean a velocity > 100 mm/s ... do you understand this can't happen?

maybe you should post the entire problem as it was given to you.

**calltronics** · August 6th 2012, 02:01 PM

I am not very good at this am I?

Travelling in a straight line between Point A and Point B

Starting at Point A at an initial velocity of 100 mm/s I have to get to Point B which is at final velocity of 40mm/s.
The movement has to be achieved in 2 seconds and over the distance of 300mm

To achieve these distance and time constraints I have to perform an acceleration/deceleration along the straight line path between Point A and Point B: -
Accelerate from Point A to a point (C) somewhere between point A and Point B
Then decelerate from this intermediate point(C) to Point B

Such that: -
the distance travelled is 300mm
the time taken is 2 seconds
and acceleration = deceleration (and at a minimum value)

Point (C) will have a velocity greater than Point A thereby basically increasing the distance travelled in the time available.

Is this any better?

**skeeter** · August 6th 2012, 03:39 PM

one more question ...

is the acceleration from 100 mm/s to vmax at C constant and equal in magnitude to the deceleration from vmax at C to 40 mm/s ?

in other words, is the velocity vs, time graph piece-wise linear ?

**calltronics** · August 7th 2012, 01:13 AM

Yes.

**ebaines** · August 7th 2012, 05:45 AM

If we let $t_1$ = the time under acceleration (and hence 2 seconds - t1 = time under deceleration), and let $d_1$ = the distance travelled under acceleration (and hence $0.3m -d_1 = d_2$ = distance travelled under deceleration) then you can set up two equations in $a$ and $t_1$ as follows. First consider the velocity at point c; under constant acceleration starting with velocity $v_a$ and then under constant deceleration to end with velocity $v_b$ you have:

$v_c = v_a + at_1 = v_b + a(T-t_1)$ where $v_c$ = peak velocity somewhere in the middle, T = total time (2 seconds)

So the first equation is

(1) $a = \frac {v_b-v_a} {2t_1-T}$

Now consider the distances travelled under acceleration from a to c and then under delleration from c to b. The total distance in the acceleration phase is $d_1 = v_a t_1 + \frac 1 2 a t_1^2$ . The distance travelled in the deceleration phase is $d_2 = v_c(T-t_1) - \frac 1 2 a (T-t_1)^2$ . Add these together, replace $v_c$ with $v_a + a t_1$ and set equal to total distance D (which is 0.3m), and after some manipulation you get your second equation:

(2) $a = \frac {D-v_aT} {2t_1T - t_1^2-\frac 1 2 T^2}$

Set equations (1) and (2) equal to each other and you get a quadratic in $t_1$ :

$(v_a - v_b)t_1^2 + 2(v_b T - D)t_1 + (TD - \frac 1 2 v_aT^2 - \frac 1 2 v_bT^2) = 0$

Apply the quadratic equation to solve, and you find that a = 0.818665m/s^2. This yields point c at 0.149m, and v_c = 0.235m/s.

**skeeter** · August 7th 2012, 06:42 AM

I saw it this way ...

let $t$ = time interval of acceleration

$2-t$ = time interval of deceleration

$a = \frac{v_c - 100}{t}$

$-a = \frac{40 - v_c}{2-t}$

solving this system for $v_c$ ...

$v_c = \frac{10(7t-10)}{t-1}$

total displacement ...

$\frac{1}{2}(100+v_c)t + \frac{1}{2}(v_c+40)(2-t) = 300$

simplifying ...

$v_c = 260 - 30t$

substituting for $v_c$ ...

$\frac{10(7t-10)}{t-1} = 260 - 30t$

simplifying ...

$3t^2 - 22t + 16 = 0$

$t \approx 0.819 \, s$

$v_c \approx 235.44 \, mm/s$

... same result as ebaines

**calltronics** · August 7th 2012, 08:03 AM

Wow!!!!!!!!!
I am so so pleased.
I got to nearly where ebains was but it took me all morning.
Trying to go back 30 years to my school maths took some doing.
Absolutely great job.
Thank you very very much both of you - amazing.
This is one part of a really complex machine movement.

**calltronics** · August 8th 2012, 12:37 PM

I am forever in your debt, but to push your kindness.......
If I now want the minimum time (T) and t1
With a fixed maximum acceleration is it easy to resolve?
Same parameters (but with variable time) and a max acceleration = 0.7m/s2 Max Accel

**ebaines** · August 8th 2012, 01:34 PM

Originally Posted by calltronics

If I now want the minimum time (T) and t1
With a fixed maximum acceleration is it easy to resolve?
Same parameters (but with variable time) and a max acceleration = 0.7m/s2 Max Accel

You will have to clarify - is the total time (T) still 2 seconds? And covering 300 mm with the same initial and final velocities as before? Because if so it can't be done, as we've already shown that a = 0.82 m/s^2.

**calltronics** · August 8th 2012, 01:38 PM

The total time is now the variable.
The acceleration is fixed at 0.7m/s^2
Same type of solution Passing through "Point C".
Solved for either t1 or t2 as a minimum total time.

**calltronics** · August 9th 2012, 03:31 AM

Originally Posted by skeeter

I saw it this way ...

let $t$ = time interval of acceleration

$2-t$ = time interval of deceleration

$a = \frac{v_c - 100}{t}$

$-a = \frac{40 - v_c}{2-t}$

solving this system for $v_c$ ...

I am lost on how you did this bit......
Can you please explain how you got to the next point?

$v_c = \frac{10(7t-10)}{t-1}$

total displacement ...

$\frac{1}{2}(100+v_c)t + \frac{1}{2}(v_c+40)(2-t) = 300$

simplifying ...

$v_c = 260 - 30t$

substituting for $v_c$ ...

$\frac{10(7t-10)}{t-1} = 260 - 30t$

simplifying ...

$3t^2 - 22t + 16 = 0$

$t \approx 0.819 \, s$

$v_c \approx 235.44 \, mm/s$

... same result as ebaines

Please see comments in your answer.

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