Second CNC table build, here we go again....

[m_c]

Smaller will give you better acceleration, and bigger higher speed.
If you know what torque you'll have, and the weight of your gantry, then you can work out your (theoretical) max acceleration. As I'm sure you'll be aware, there's no use having a high top speed if you don't have the acceleration to make use of it.

[Davek0974]

I'm now looking at direct drive, a 17t pinion will give me 53.39mm of travel so I will need 168rpm to reach my top speed of 9000mm/min.

The gantry will weigh at a guesstimate 30kg and will have two 3.1Nm motors, direct driving it.

This might work better as it keeps the motor rpm way down in the high torque area.

Any chance you could illustrate the maths needed for the acceleration estimates???

[Robin Hewitt]

I'm now looking at direct drive, a 17t pinion will give me 53.39mm of travel so I will need 168rpm to reach my top speed of 9000mm/min.

The gantry will weigh at a guesstimate 30kg and will have two 3.1Nm motors, direct driving it.

This might work better as it keeps the motor rpm way down in the high torque area.

Any chance you could illustrate the maths needed for the acceleration estimates???

2pi/.05339 gives you a 117.68 advantage over a 1m pulley

So 2 x 3.1 x 117.68 = 730N

So a 30kg gantry will accelerate at 730/30 = 24 m/s/s about 2.4G on a good day with a following wind.

Assuming I haven't totally cocked up, someone should check this

[Davek0974]

Thanks, if that is indeed correct then I have slightly more acceleration than needed, even allowing lavishly for other losses:)

I have read that plasma cutting only needs around 0.2 to 0.5G for quality so it does sound encouraging.

[Robin Hewitt]

So what is the bending force on your gantry from accelerating the Z axis components?
Suppose that is 3kg of motor, slide, torch, screws and stuff. Wild over estimate.
We know that 1kg force will accelerate 1kg mass at 1G because that is what happens when you drop it.
So to accelerate 3Kg mass at 0.5G requires a force of 1.5kgf
Why do you need 30kg of gantry to withstand 1.5kg in X and Y?
Perhaps it is the 3kg in Z you are worrying about?
Perhaps you should take a few gantry length bars, support them at either end, put a weight in the middle and see how far they actually bend.
Maybe 30kg is not necessary after all?
Maybe 30kg is only there to accelerate the 27kg it contributes all by itself?

[Davek0974]

Ideally, it needs a structure that resists torsion moments - this is where my current machine fails - the gantry is a length of 44x44 structural aluminium which is ok in the beam but poor in tension as the centre box is only 18mmx18mm.

The torch and axis are not balanced and when accelerating it twists one way and the opposite on decelerating - causing ripples in the cuts.

What is ideal is a simple, light structure that ca withstand torsion or twist as well as support 3-4kg in the centre.

What that is, I have no idea, most machines just use box section because it's there I guess.

It needs a thick enough wall section to take 5-6mm screws firmly so I guess 5-6mm thick.

Without getting a structural engineer on the job, what stock item would be best suited here?

[Robin Hewitt]

When I started making one I got a piece of 4"x4"x10g aluminium box and made the gantry as the long axis. Across the front I bolted a linear rail, using the bolt from behind kind so the dust could not get inside the block via the bolt holes. I used 4 1/2" bolts and spacers so I could bolt everything from behind without crushing the box. I cut the box ends dead square. I inserted 1/2" square plates with 8mm steel nut inserts dead centre from either end. The gantry end plates had 4" x 4" recesses and an 8mm bolt hole. When I tightened that sucker up it became incredibly rigid. The gantry end plates got double round linear blocks running on 30mm round linear rails which held the assembly in place simply by being bloody heavy. I liked my design but it all went horribly wrong because the 5mm pulleys were such a crap fit on the 5mm belt. I lost heart.