Force needed for tilt/pan

IVAN__V

Hi people.
I'm trying to figure out how to build a rotating platform thats powered by an electric dc motor. Basically i want to spin this platform say 1 m in diameter, and it's got an object on top of it. Like say anywhere from 5 kg to 50 kg or even maybe 100 kg. What i know for sure is that i'm interested in using a worm wheel type gearing. I've looked around at dc motors and power in the 100's watt/kw range...
The problem is i just don't know how big a motor or should i say powerfull i need to use , before i go out and buy one and realize it's too small/weak.
Thing is i need it to rotate the platform both ways at a max 60 rpm. but it needs to be able to switch direction fast....thats why i'm planning on using a worm gear ...to be able to gain a lot of leverage and i also have to option of moving it slower and like have a very good degree of precision.
So how do i calculate the watts needed... if i go with a 1500rpm motor times a 25 reduction worm gear..that gives me 60 rpm, so that i can change it's direction ...i want it to be snappy , like resemble a pan/tilt thingy used for cameras.

Edit.
So like 0.5...1...1.5m diameter (let's consider the weight on it sort of well balanced for simplicity)
5...50..100 kg weight
dc motor initial rpm doesn't really matter cause i can always find a proper worm drive to get it down to 60 rpm.

IVAN__V

I'd really appreciate someone's help.
Ok. I'll try and make it simpler:

How much energy , in N/m or some convertible unit, a disk has if it's spinning at 60 rpm and it's 1 meter in diameter and weighs 50 kg.
Maybe someone knows a formula to calculate the energy for different values of it's weight, rpm, diameter.

IVAN__V

Ok. I've managed to find something out..i'm not sure if it's the right path.

I've found that using rpm P(power in Watt) = T(torque) *2*3.1415 * rps =
so P=T*6.283*1 => P=T*6.283
and if i want to know T then T=P/6.283

Does this look right ? So the T torque of a 1000 watt motor is T=1000/6.283 ? of course given the 60 rpm or 1 rps speed of my motor.

So this takes care of figuring out the motor torque.

But how about the spinning disk torque ?
I've found two equations but they describe a weight at the end of a bar that's rotating. Like a horizontal pendulum.
T=r*F torque=radius *Force
and
L=r*p Linear momentum=radius*radial momentum

The thing is the first formula is for a system thats like a half disk, it isn't balanced on the horizontal plane. Mine is a disk. Also i'm not sure if i even need or for what the second formula is useful.

IVAN__V

I've also found this:
F(force)=m(mass)*a(acceleration)
and for circular motion, my case, right ?
it's a(acceleration)=v^2(speed squared)/r(radius)
Now v at the edge of 1m disk is 3.14 m/s and r=0.5
so a=3.14^2/0.5=19.72
now since F=m*a then F=50kg*19.72 so F=986

Is that correct ? 986 Newton meter ? or what ?

tygerdawg

You're approaching this problem backwards.

You must do the dynamic motion analysis to determine the peak torque required to move the turntable+load in all loading conditions. Your motor+gearing must be sized for that, not specified up front and then forced to work. That's a sure road to failure.

T-peak = T-accels + T-steadystate + T-friction + T-gravity + T-safetyfactor + T-gearingefficiency + T-phaseofmoon + T-this + T-that + T-theotherthing

Calculated for starting and/or stopping and any other conditions that you could possibly conceive of.

From this peak torque value you can calculate power required (watts or horsepower) and then try to find a suitable gearmotor combination. Motor must be sized for number of starts per hour (for heat dissipation) and final speed required. Then you must select a suitable controller to be able to drive the motor to required speeds.

Websearch for "Smart Motion Cheat Sheet" in PDF format, it has all of the essential dyanmics equations.

Gearmotor manufacturers such as SEW-Eurodrive, Dodge, Nord, others ususally have useful & freely-downloadable engineering guides where you can learn how to calculate this stuff (in case you failed your Dynamics class).