Robot DC motors

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fortion

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i am making a robot with a rotating disk of 8KG and diameter of 1.5 ft . i want to rotate this disc using DC gear motor at a speed of 100 RPM . by my calculation that would require a an output power of 30 W . can someone confirm this calculation ??? it looks wrong to me . also if i use four 30W gear motors of different RPMs namely 100 , 200 , 300 , 400 and use them simultaneously to run the disc will the disc achieve 400 RPM ??? will this assembly work ??? i am building this robot as a hobby to compete in robo war on a local level . if anyone can help i will appreciate it
 
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oh and also i am using 12 V 5 Ah DC batteries to run this assembly . if anyone can help i will appreciate it
 
Are you trying to make a gyroscope (flywheel) to stabilize the robot?
 
first of all i calculated I , moment of inertia , 1/2m(r square) . Then i calculated torque which is torque= I*(rad/sec) . I put 100 RPM after converting it . and then finally Power = torque*(rad/sec) . is this method correct ???
 
is this method correct ?
Click to expand...
Only partly. I and torque enable angular acceleration to be calculated. You also need a time factor to get rpm. Your Watts will determine acceleration, not rpm. In theory, in a lossless system, no power would be used to maintain a given rpm (but of course your flywheel won't be lossless ).
 
thanks a lot . i see my mistake now
 
ok so this time i followed this method :-

l i calculated I , moment of inertia , 1/2m(r square) . i want my disc to rotate at 100RPM (10.466 rad/sec ) and want it to attain this speed in 2 secs . so i calculated (Angular Acceleration)=(rad/sec)/time . next i calculated torque by torque= I*angular acceleration . and finally i used POWER = torque * (rad/sec) . the answer comes out to be 11.44 W . i did all this calculation in SI units . is my calculation correct now ?
 
i used mass 8 kg to calculate the moment of inertia . i used the formula I= 1/2 m*r squared . i think this is correct
 
if it's a weapon, forget about a gearmotor; use a series/universal motor give it a couple minutes to spin up, and let 'er rip. Can you imagine that disk spiining @ 20,000 RPM when it hits your foe? Instant win.
 
alec_t said:
Well I got 16.4W using an online units converter. So one of us is a bit out .
Click to expand...

did that convertor calculate output or input power . did you multiply it with a service factor ?? if you calculated output then you are probably right because i was converting units by myself so possibly i made a mistake . i will recheck
 
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strantor said:
if it's a weapon, forget about a gearmotor; use a series/universal motor give it a couple minutes to spin up, and let 'er rip. Can you imagine that disk spiining @ 20,000 RPM when it hits your foe? Instant win.
Click to expand...

if i use a series motor it will give my weapon very good RPM no doubt but would it be able to give me a continuous good torque . i mean if i use a geared universal motor , i.e limit its rpm but increase its torque , would it be more effective or not ??
 
did that convertor calculate output or input power
Click to expand...
Neither. It converted a calculated energy value in Kg.metres to Watt.secs
did you multiply it with a service factor ?
Click to expand...
No. What is a 'service factor' in this context ? What do you regard as 'output' here? I was calculating the theoretical power (in Watts) required to accelerate an 8kg disc in a lossless system. I have a feeling I may have lost a factor of 2xpi somewhere, though. 16W seems far too low a figure. My applied maths is pretty rusty . Anyone else out there feel like doing the sums?
 
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alec_t said:
What do you regard as 'output' here?
Click to expand...
by output i mean that when we are in a system which is not loss less , there will be some difference between input power supplied to the motor and the output power given by the motor . so the input power will be greater by a factor than output power
 
okay redid the calculation this time using online convertors :

m=8 kg

dia.=1.5 ft

so , r=.75 ft = .2286m

I = 1/2 m*(r squared) = (0.5)(8)(.2286)square
=(4)(0.05225)= .20903184

ω=100 RPM =10.4719755 rad/s

t=2 sec

α(angular acceleration) =ω/t = 10.4719755/2 = 5.235985 rad/sec squared

τ(torque)= I * α = .20903184 * 5.235985 = 1.094487579 Nm

P (power) = τ * ω = 1.094487579 * 10.4719755 = 11.46144 W

is this method alright ??
 
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