distance = rate * time

I dare to ask.
acceleration and deceleration on top of this simple formula.

Was sent this.

flew over my head.

V = D/T (Average Velocity = Change in Position /Change in Time)

A = V/T (Average Acceleration = Change in Velocity/ Change in Time (ft/sec-sec)
V = A*T (Average Acceleration * Time)

A = D/(T squared)

D = ((initial velocity + final velocity)  2) * T (D = Average Velocity * Time)
D = ½ * A*T squared (This applies only when object is accelerated from rest)
D = (initial velocity * T) + ½ A* T squared (Distance traveled while accelerating for a given time)

From above:
Distance traveled decelerating to a stop is: D = V/2 * T & therefore T = 2*D/V

T = 2*D/V
is only correct under these assumtions:
V is actually final velocity.
Initial velocity is zero.
Acceleration is constant.

A more gereric formula would be:
T = D/Vav
Where Vav is the "average" velocity.

edit:
In real life it is easy to measure time spent and distance travelled, but more difficult to measure velocity and accelleration.
If trying to derive a position from measured velocity and accelleration, then there will be inaccuracies.

If I read the motor frequency back from the drive, that should get me closer, correct?

???

Why did I even respond to the 1st post ? Well that wont happen again.

JesperMP said:

???

Why did I even respond to the 1st post ? Well that wont happen again.

Click to expand...

JesperMP said:

???

Why did I even respond to the 1st post ? Well that wont happen again.

Click to expand...

well, the original programmer walked off the job and said it couldn't be done.

Seeing how the customer hugged me today, maybe I'm not an idiot after all.