Hi all

[8iggles]

Interesting bit of construction.. I take it thats your machine minus its drive trains.

re belt drive, I can't help but feel there's going to be some difficulty maintaining accurate postioning with an open belt. What pitch are you thinking of?

The tutorial is here

Yes no drive trains yet, as I said in 1st post budget..the lack of spare cash - I've used threaded rod for cost and adjustment

The belts I have are 5mm pitch and pinion is 10 teeth then reduction before stepper motor, I think the change in direction will be the largest force on the motor.
Its still a blank canvas so I can go for some decent ballscrews if all else fails, but again it all comes back to cost and failiure is not an option

I'll try and get some pics of where i'm at with the drive

[irving2008]

Yes no drive trains yet, as I said in 1st post budget..the lack of spare cash - I've used threaded rod for cost and adjustment

The belts I have are 5mm pitch and pinion is 10 teeth then reduction before stepper motor, I think the change in direction will be the largest force on the motor.
Its still a blank canvas so I can go for some decent ballscrews if all else fails, but again it all comes back to cost and failiure is not an option

I'll try and get some pics of where i'm at with the drive

I have done some thinking and reading around timing pulleys and belts. It all comes down to the shear force on the teeth. The load has three components - static force, cutting force and inertia.

The static force is given by F = m * u where m is mass and u is friction coeff.. for ball races say 0.01, and lets say gantry is 30kg so F = 0.3N.
Cutting force for most woods would be 5N. Inertia is given by F = m * a. we want to accelerate to maximum speed in 0.02sec (for a medium spec machine). maximum speed for cutting is, say, 1200 mm/min = 0.02m/s. The acceleration therefore is .02m/s divided by 0.02 sec = 1m/sec^2 so F = 30 * 1 = 30N

so our pulley has to push on the belt with a force of 0.3 + 5 + 30 = 35N (ignore the 0.3, its noise!).

At 1200mm/min the pulley revs, for a 10 tooth pulley on 5mm pitch will be 1200/(5 * 10) = 24rpm = 0.4rps (2.5rads/sec)

The diameter of a 10tooth pulley on a 5mm HTD is 15mm so the radius is 7.5mm and the torque therefore is 35N @ 7.5mm radius = 0.26Nm

Power = Torque * rads/sec = 0.26 * 2.5 = 0.65W

From ther HPCGears catalog pages 787-790 the required HTD belt can be calculated as follows. Choose service factor c0 as 2.0, calculate c2 = c0 + c3 + c6 = 2.0, taking C3 and C6 both = 0. Pb = Power x C2 =.65 x 2 = 1.3W, which means from Graph 1 (page 788) aN 8M belt is needed.
there is one tooth in mesh so C1 factor is 0.1 and C7 taken as 0.8, therefore Pu > Pb and Pu = Pn * c1 * c7 so Pn > pb/(c1 * c7) = 1.3/(.1 * .8) = 16.25W which suggests a 15mm wide belt minimum. Note however that the torque load of a 15mm dia pully is a bit high.

A better choice would be a 20 tooth pulley, with a OD of 32mm, radius 16mm, torque 35N @ 16mm = 0.56Nm and Pb = 0.56 * 1.25rads/sec^2 = 0.67W so Pn>16W as before. This also needs a 15mm wide belt but is more in the sweet spot in the belt tables.

Now, 1 rev of pulley = 100mm movement, and assuming we want 0.05mm resolution we need 1 rev of motor = 10mm so we need 10:1 gearing, so motor speed at 1200mm/min = 2 revs/sec or 400 steps/sec.

So motor needs 10tooth, driving 100tooth/20tooth compound gear driving against the belt. Mechanically could be tricky to arrange.

On this basis only a tiny motor would be required due to the gearing... It doesn't look quite right to me, I need to think it through more...

[8iggles]

More pics...







Wow more homework, thanks so much for the info...I'm using 2 rollers so theres at least 5 teeth in mesh and as luck would have it I got 15mm wide belt for the X axis and 10mm for the Y axis...I was worried about the inertia when changing speed / direction of the gantry - it runs freely but is quite heavy

The small router on the first pics is only a temp thing just to let off some of the excitement that's building and demo it to my son - I'm hoping to make a bearing block and end up with a keyless chuck driven down the centre of the square tube from a motor on top...I thought it would be better in the centre of the bearings rather than overhung to one side ??

Thanks again
Shaun