Thread: Plasma Attempt #2

19052012 #1
Me again
First plasma attempt a dismal failure, next attempt is a pile of bits.
A new, slightly bizzarre design which seems to work on paper. I'm typing this because explaining the logic behind it will force me to check my maths...
What happens when you put 10kg bag of lead shot in the middle of a 3 meter aluminium box section with a DTI underneath?
Answer: It bends down approx 0.5mm. Not bad for something that only weighs 10kg itself !
"Hang on twit brains", I hear you cry, "the gantry is usually the short axis".
Possibly so, but going the wrong way round does have advantages when it comes to positioning the machine and loading metal on to it. Also Gary managed to send me a 2.7m length of HiWin rail unbent and it would seem ungrateful not to give it a go after that.
Here's the maths...
Let me aim for a tight plasma cutting tolerance, say 0.1 mm of gantry flex allowed. Using Hookes Law, if 10 kgf bends it 0.5mm, 2 kgf will bend it the allowable 0.1 mm. Simples.
The whole gantry weighs 20 kg but only about 6 kg of that contributes to bending because the stuff at the ends doesn't count.
Obviously 6 kgf will accelerates 6 kg at 1G, that's what happens if you drop it. So it follows that my 2 kgf will accelerate it at 1/3 G. Maths is easy if you choose the right units. Tom Caudle reckons you need 0.1 G accelerations for plasma so I am well inside the ball park.
To accelerate the whole 20 kg lump at 1/3 G will require about 20 * 1/3 = 7 kgf, call it 70 Newtons.
I like to cut in mm and a stepper motor does 200 steps per rev. To keep the units sweet 20 mm travel/motor rev suggests itself, but that is dismally slow given that steppers tend to clack out beyond 5 rps. I think I prefer 40 mm travel/motor rev. That gives me end to end on a 2.5m sheet in 12.5 seconds. Slow, but not dismally slow.
6.28 divided by my 0.04m/rev gives the motor a mechanical advantage of 157:1
So if I want 70 Newtons to accelerate it, I need a motor which can deliver 70/157 = 0.44 Nm. Double that for luck and a 0.8 Nm motor should do the trick. Seems pathetically small for such a huge table but you can't argue with Mr Newton.
First job is to drill the box section to mount the HiWin rail which is the threaded from below variety. The plan is to run M5x110mm bolts in from the back with spacers in the middle. It is most important that I can get at the bolt heads because I intend to bend the rail slightly up in the middle so the torch flies at a constant height.
Specially for Jazz I'm using unsupported round rail for the Y axis. The Z axis is truly weird, took me the best part of a week to figure it out
best
Robin

19052012 #2
Where's the maths, I can only see arithmetic?
You should include the moment of inertia of the pulleys in the calculation since especially with a large reduction it will contribute significantly to the overall torque requirements.
It looks like the SY57STH763008B motor from Zapp is appropriate since at 300rpm the torque output will be about 1.3Nm on 50V drivers. However considering that motor is more expensive than the 3Nm motors at CNC4YOU and they will develop 1.3Nm at 1300rpm on 70V enabling you to either speed up or increase resolution I know which I'd go for...
40mm per revolution is 0.2mm/step  not a good starting point for a 'tight plasma cutting tolerance' of 0.1mm.
What diameter and length is this unsupported rail on the Yaxis?

19052012 #3
I passed the 11+ and went to a grammar school where they taught us maths, arithmetic was for those plebs down at the secondary modern
The motors drive 3:1 down on to 24 tooth T5 open belts. I got some 15mm wide, steel reinforced belting which is probably massive overkill but I'd rather it stretched as little as possible. Nice and light to.
There is a 2.5m axle across the back, I think 1.5" 10g aluminium tubing should be fine for that. At <2rps it should not get the shakes .
The unsupported rails are 30mm x 1600mm with the longer LUU type bearing blocks, one at either end. There are no cutting forces to contend with so they only need to support 10 kg each. The bearings are kind of chunky at about 1.25 kg each, but I think that is mass well spent.

19052012 #4
That's to link the belt drive I assume? If so the inertia of that tube will be immense; I=pi/2*7850*2.5*((38.1/2/1000)^4(31.3/2/1000)^4)=0.0022kgm^2. To put that into perspective that's only half the moment of inertia of a 1.5" solid bar.
You'd probably be better off with bar. For 2rps critical speed and 2500mm length that's only 5mm diameter, so use anything between 5mm and 33mm solid bar and and have less inertia than the tube yet still have high enough critical speed. Tube will have higher torsional stiffness so could do with comparing that.

19052012 #5
Forgot to say, 0.0022kgm^2 will add about 0.45Nm to the stepper torque requirement  angular acceleration of shaft is 4*pig/0.6. So torque=4*pig/0.6*0.0022=0.45Nm. So there goes your 2* factor of safety!
If instead of the tube you used 12mm steel bar it's only 0.0082Nm added, which is about the same torque as a 24T HTD pulley will require. Critical speed of 12mm bar is fine, but to avoid needless vibration you could pop a bearing in the centre to support it.
Those powers of 4 really do give some surprising results.Last edited by Jonathan; 19052012 at 10:04 PM.

19052012 #6
Good man, you are making me justify it, but I do think you are making the maths slightly complicated...
The 1.5" diameter tube has a circumference of 120 mm, same as the PD of the pulley. So in 1 turn it moves the same distance as the gantry. I can simply add the tube weight to the gantry weight.
OTOH, point taken, 16 gauge can handle 0.66 Nm of torque, I can save myself 1.4kg
Let's up the gantry weight estimate from 20 to 25 kg, required torque goes up in proportion 0.44 Nm becomes 0.55 Nm. To get the safety margin I either need a bigger motor or a lower acceleration.

19052012 #7
Wait what... I thought this tube was spinning? Either way yes you do need to add its weight to the gantry, but if it's spinning the moment of inertia of the tube is far more significant than that.
Moment of inertia of 16 gauge is 3.35 times less than 10 gauge, so 0.135Nm as opposed to 0.45Nm  still a lot.
P.S. By my definition I've not used any maths yet, let alone made it slightly complicated!

20052012 #8
This is the difference between maths and arithmetic. I happily put all the mass on the outside diameter, assume 10g is twice the weight of 16g and overestimate the inertia somewhat.
You work to three decimal places and we disagree
Incidentally, what thickness are you using for gauge? Once you get past tiny the OD goes up in multiples of 1/8". People like tubing to fit snug one inside another so the wall thickness is usually 1/16" for 16g and 1/8" for 10g.
My 4 x 4 * 10g box section has a 1/8" wall, I just measured it.

20052012 #9
Nah for a skinny plasma with 30mm rails I'll let you have this one.!! . . . . . But all this Math or Arithmatic shit is making my crust hurt so just stick a bloody 3Nm motor on plug into national grid and watch the thing go like a rocket.!! . . . Don't matter all corners will all be circles and the thing will rip it's self apart trying to speed up/slow down and change direction unless you have a nice Scurve motion planner..

21052012 #10
Not quite  we disagreed because we're talking about fundamentally different things. I should have explained myself more clearly. It's three significant figures anyway :whistling:
To accelerate the gantry the motor needs to impart translational kinetic energy into the whole gantry and rotational kinetic energy into the rotating parts. Therefore we must consider the sum of the torque contribution from each. So far you've only considered the translational kinetic energy, so I worked out the torque contribution from rotation. For that instead of calculating the inertia, you find the moment of inertia (I), which is a measure of an objects resistance to changes in its angular velocity and use the formula torque=angular acceleration * moment of inertia, where acceleration is in rad/s^2.
In post #4 I calculated the moment of inertia of your 10g (well not quite, see below) tube using the formula, courtesy of Wikipedia:
In post #5 I calculated the acceleration of the tube and stuck that in the above formula for torque to get 0.45Nm and subsequently 0.135Nm. With the mass of the tube included your estimate for the torque due linear motion is now 0.55Nm, so just add that to the figures I get to find the overall system, i.e. 1.0Nm or 0.68Nm. Or if you used 12mm steel bar it's <0.01Nm added... I should also calculate the moment of inertia of each pulley and add that to the system, but it's not going to change the conclusion. Either way the 3Nm motors even on 50V drivers (although clearly 70V preferable) will be a good match and surely the best option since they are the best price/Nm!
I used:
http://www.engineeringtoolbox.com/ga...eetd_915.html
Is this some silly American/English difference in the gauge system (like with wire gauges)?
It would be interesting to know how much it actually stretches  i.e. find out the spring constant. You could hang up a length of it and attach something heavy, put DTI underneath and zero it, then add known mass and see by how much it stretches. I've spent ages trying to find the spring constant of different types of timing belts, but just can't find anything.
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