Wich motor tot use with 6dof

Wich motor should I use when I want to make a 6dof with a platform of around 300kg?

Going safe, you want a motor to lift at least 300kg.
With a linear actuator using a SFU1605 ball screw (5mm per turn with 16mm diameter), you need a motor with:

Nm=300kg*8mm/1000mm
Nm=2.4

I'm suspect, but I would go with something brushless.
For high precision, under 0,2mm I would use an encoder.
Lower resolution, I would use a system like I'm using on my system. Reading the bldc hall sensors.

Hey there,.. :-D

Here's the line or reasoning that eventually lead to this post I made on a similar topic. As always, I'm open for discussion.

- W = Moving Weight [KG]= Platform weight + Payload.
- If you wanna move W KG (300Kg is a LOT!) then you need to carry a total load of ~10W in Newtons [N].
- So each actuator needs to carry 10W/6 of vertical load in Newtons [N].
- In the most unfavourable constellation that I still want to be covered (60° Angle from vertical --> 1/cos60° = 2) the actuator needs to carry 2*10W/6 [N] of load along it's axis. That is only to carry the 1G static load! No vertical motion!
- Now, if you wanna accelerate the platform & payload at an additional 1 G (my minimum design specs) then you need your actuator to produce 2*2*10W/6[N] = 6,6666*W[N]. In your case: 6,666 * 300 = 2000N of Force per actuator

...that's how I came up with Rule N° 3 in my rules of thumb in the above post. For simplicity: „Min force = 6*W [N]“

- If your actuator needs to produce ~6*W of force [N] with a ballscrew, then you will need your motor to have a Torque[NM] of at least Pitch[mm] * MovingWeight[kg] / 1050. In your case: 5[mm]*300[Kg]/1050 = 1,42 [NM]. That’s the torque your motor needs to accelerate the platform up at 1G (…on top of the natural 1G).

The only question that remains is now: How fast will will the stroke of your actuators be when under such load? Take a look at which RPM the Motor produces its max torque. Im my case it produces 2,8NM (peak) until ~3700PRM...


...so I could expect to reach an…

Actuation speed [mm/s] = RPM * Pitch[mm] /60
=> 3700[RPM]*5[mm]/60 = 308[mm/s]

…under max load, in the most unfavourable constellation! But that is only the actuator speed! Because of the geometric constellation, that will mean that the platform itself will move up at 616mm/s (60° Angle from vertical --> 1/cos60° = 2).

Bottom line:

• Take 6 Motors of 1000W (peak) each.
• Make sure they deliver 1,5NM of torque…
• …at about 2500RPM or more.
• Be prepared for a „2G-experience“ during operation

…and really make sure wether it is absolutely necessary for your moving weight to be 300Kg. That’s like two persons with a cockpit!?!

Dirty

oh,... and don't post the same question twice

Hey :-D

...is it correct to use the half diameter (8mm) there? Because it should depend only on the pitch. Why would a ballscrew with a bigger diameter require a higher torque? It's still one revolution and it moves the rod by Pitch[mm] forward.

Please correct me if I'm wrong.

Dirty

You are right...
Well I have to start thinking before writing. Don't know what I was thinking...

Only thing wrong is that pitch is different from lead.
We should use lead. That's the travel for one turn, while pitch is the distance between threads.
Usually the same in common ball screws, but pitch can be half the lead if it's a "double screw" (with two threads).

Making my own calculus:

Motor_Torque = Tangential_Force * Diameter/2

But the force is the tangential one, not axial.
To calculate the axial force we have to consider the lead of the ball screw.
So for one lead we have one perimeter:

Lead --> Perimeter = PI x Diameter

So forces are proportional. Solving, we have:
Axial_Force / Tangential_Force = PI * Diameter / Lead

Replacing tangential force:
Axial_Force / (2 * Motor_Torque / Diameter) = PI * Diameter / Lead

Axial_Force(N) = 2 * Motor_Torque(Nm) * PI / Lead(m)

Or:

Motor_Torque(Nm) = Axial_Force(N) * Lead(m) / (2 * PI)

Don't forget that 1kg = 9,81 N and 1 mm = 0,001 m
We have to use the same units everywhere.

The travel speed of the actuator will be given by:

Travel_Speed(mm/s) = Lead(mm) * RPM / 60

So for each actuator, you need 300/6 kg = 300*9,81/6 N = 490,5 N
Like Dirty says, for maybe the worst situation (60º) and 2 G, multiply this by 2 x 2:
490,5 * 2 * 2 = 1962 N
This is axial force. Motor torque for a 5 mm lead ball screw should be:
Motor_Torque = 1962 * 0,005 / (2 * PI) = 1,56 Nm

All this is ignoring friction and other factors.
But you are safe with this.
Just get a motor with a rated torque and speed above the calculated values.

In my case I have a rated value of 0,6 Nm and 3500 rpm with load.
Peak values are 1,8 Nm and 4500 rpm without load.
This is a safe margin in my opinion.
With those values I have an axial force of 76,85 kg at rated power and 230 kg at peak.
By Dirty calculus, my platform should not have more than 114 kg.... I have to loose some weight
But there's always the peak power to help. In the end, I might need another power supply to allow full power to all actuators at the same time and get to 345 kg.

Always design to safe and rated values.
In practice, and in my opinion, for car sims, we use small travels and higher frequencies, while for air planes we use larger travels, and lower frequencies. So the 60º might not be reached frequently.

I fully agree!!!

It's good to see that you came to identical results, really glad to read that. And thanks for the tip about Lead vs. Pitch. That never would've occurred to me.

Waldo,
60° from the vertical will indeed almost never be reached, 2G will almost never be required, peak power will almost never be needed. But still, you want to establish SOME limits to be sure that you are on the safe side of whatever you specified.

If you build something that costs a few months worth of salary and takes ~2 years to built,... at the very least you want to screw up on purpose, not by coincidence You can change those specifications at any time if you deem it necessary, but you should not build without any.

Maybe this Excel sheet can help you.

Dirty