I used Excel to help me create a script for the rotating object task you posted. You have 4 events in the rotation.
1. Initial stationary frame (I assume F = 0).
2. Frame at which rotation reaches constant angular velocity ( I assume F=25)
3. Frame at which rotation starts to slow down (F= 85)
4. Frame at which rotation stops (I'll assume an instantaneous stop for this example.)
If the rpm = 100 and there are 30 FPS then the angle displacement per frame after the constant speeed is reached is 20°/sec. If we assume a gradual speeding up of rotation from rest then we can say:
angle = a * F^2 where a is the coefficient needed to get to a set angle at the specified frame.
If the shaft rotated from rest at a constant velocity it would reach an angle of 600° by frame 25. SInce we want to start slowly we can set a goal of something less than 600 for frame 25. I'll assume 300°.
Therefore:
a = 300/25^2 = 0.48
So, for the first 25 frames we have
angle = 0.48 * F^2
From frame 25 to 85 there's constant speed so the angle increase linearly with the slope of the curve equal to the speed.
After frame 85 the angle remains constant.
The equation gets a little tricky to slow the rotation down graually so for now I will just assume an abrupt stop as you can see in the attached file.
Here's the rotation angle graphed in Excel.

The script in Max. I've included some format statements to help in the debugging process. They of can be commented out or deleted.

degPerSec = rpm *360 / (60 * 30)
format "\nDeg/sec = %" degPerSec
a = 300.0 / 25^2
format "\na = %" a
if (F <= 25) then
(ang = degToRad(a * F^2)
)
else
(
if (F <= 85) then
(ang = degToRad((F - 25) * degPerSec + 300)
)
else
(ang = degToRad(60 * degPerSec + 300)
)
)
format "\n\nF = %, ang = %" F ang
ang
As for your statement about "negative" rotation, there is no real positive or negative rotation! The perception of rotation is a function of the perceived difference from one frame to another. If the teapot were initially pointing straight up (12 o'clock) then rotated to a position of 11 o'clock (i.e., 30° CCW) you would perceive this to be a CCW rotation. However, if the teapot were rotate by 330° CCW thus pointing to 1 o'clock you would perceive it as a clockwise rotation of 30°. Which is it? A 30° CW rotation or a CCW rotation of 330°? Your eyes are more comfortable seeing it as a CW rotation. The wheels on stage coaches in old western movies sometimes appear to be rotating backwards because of this effect.
lee.minardi