OK you’ve specified an acceleration of 1m/sec/sec as an acceptable acceleration. I’ll also interpret that as a 5 sec time, 0 to max speed. I do this because the torque vs RPM curve of a stepper isn’t a flat line, a constant number. It starts out “high” and decreases w/RPM. Like a car, your platform will accelerate more quickly at low speeds vs higher speeds.
Your relation btw platform linear speed and RPM was almost right. I assume there’s a gear on the stepper to engage a gear on the drive pulley. Most likely (or not, see below) this will reduce the RPM from shaft to pulley, but it will increase the torque (by the inverse ratio).
So I would approach the question of motor requirements thusly;
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It’s got to have enough torque at max RPM to overcome the resistance to motion your platform will have at max speed. The pulley size (where it contacts the line) and the above gear ratio play into this.
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The integrated torque curve must be great enough to get your 0 - 5m/s in 5 secs. The pulley size, gear ratio, platform mass and resistance all play into this along w/the torque curve. This may be the hardest one to get a handle on.
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The max RPM, accounting for gear ratio and pulley size, must get you your 5m/sec max speed.
Let me assume your 30 mm pulley has a 30 mm diameter. To go 5m/sec means it has to revolve 5000/(30 x pi) = 53 times in 1 sec. That’s (x 60) 3183 RPM. So with a 3600 RPM motor you could run > that speed with 1:1 gear ratio, assuming the motor had enough torque to overcome whatever resistance there was. So what might that resistance be ?
Frankly I don’t know. I do know that you could measure it fairly easily if you have your platform built and a tramline setup. For the moment let me ASSume it’s 0.1 kgf (only because it makes the math easy). Let me also assume a safety factor of 2, meaning the motor, through gearing, must produce at least 2x that retarding force. So 0.2 kgf at the radius of your pulley (30/2 = 15 mm = 1.5 cm) is .2/1.5 kg cm = 0.14 kg cm @ ~3200 RPM. Alas you’ll have to find a very complete datasheet to get that sort of number for a motor. FWIW 1 kgf = 9.8N, as in Newtons of force, the proper SI unit.
Your 1’st link to an Ebay stepper would get you a motor roughly equivalent to this one;
https://www.sparkfun.com/products/10846
From the manufacturers site I guesstimate it makes 20 N cm at 1000 RPM and a peak of 43 N cm at 175 RPM. That 43 N cm is close enough (= 61 oz in) to the 68 oz in claimed that I’ll call it the same. So that’s a min torque (in non-SI units) of 20/9.8 = ~2.0 kgf cm. But it’s a 1000 RPM motor and you need about 3x that. So the 3:1 gear ratio means that 2 is only 0.66 kgf cm at the pulley. More than enough over the 0.14 kgf cm, assuming the SWAG for resistance/retarding force is correct.
So what kind of acceleration might you expect ? Let’s model the torque curve (vs RPM) as a right triangle on top of a box, the height (torque) of the box being the min torque of 20 N cm. The height of the triangle is then 43 - 20 = 23 N cm over a base of 1000 RPM. A box of equivalent area with an equal base would have a height (torque) of 1/2 the height of 23, or about 11.5 N cm, which when combined with the min torque of 20 N cm means, on average, across the entire 1000 RPM, the motor will produce 33.5 N cm of torque. That means a force of X at the pulley groove, X = 33.5 N cm / 9.8 N/kgf / 3(gear ratio) / 1.5 cm (pulley radius) = 0.76 kgf. Subtract the ASSUMED resistance of 0.1 kgf and you’ve got a net force of ~ 0.66 kgf to accelerate your 3 kg of platform. The definition of 1 N of force is that it will accelerate 1 kg of mass at 1 m/sec/sec. Since 0.66 kgf = 6.53 N, you should have more than enough torque, on average, to meet your accel requirement, ASSUMING the guess for retarding force is anywhere near correct. Also I’ve neglected the moments of all the rotating bits. That may or may not be a good assumption but expect real life acceleration to be reduced.
So far it appears on casual examination that the SF, or Ebay (??), stepper might work.
Your last question asked if the same motor could do 0.01 m/sec. I don’t know. Obviously you could step, stop ‘n’ wait, step, stop ‘n’ wait, … to achieve any slow average speed, but I don’t think that’s what you want (unless it’s time lapsed w/motion photography, speeded up to make a ‘movie’ … in which case it’ll do). A step is typically 1.8 deg or ~1.5 mm of linear travel w/your pulley and 3:1 gearing. Given microstepping can maybe give you 10x finer resolution, I don’t know at what speed such stepping becomes too variable in size and timing and also becomes stop ‘n’ wait jerky. Perhaps someone else can answer.