I mean two jack shafts with the belt fixed to the gantry and the servo 
driving a jack shaft.

JT

On 9/11/2015 1:40 AM, Milosz K. wrote:
> If you mean by an open ended timing belt that is fixed at both ends and the
> motor is traversing along the belt then, yes. One glaring omission in my
> initial reply was not accounting for a speed reduction. I'm guess there
> will be one between your servo and timing belt pulley. In that case you
> have to account for speed reduction:
>
> w/N
> alpha/N
> Jreflected = ((gantry m + motor m) * R^2) /N^2
> Jt=Jreflected + Jmotor + Jreducer
> Trl=(Ffriction * R)/(n*N)
>
> 2015-09-10 9:55 GMT-04:00 Dave Cole <[email protected]>:
>
>> Yes.
>>
>> On 9/10/2015 7:36 AM, John Thornton wrote:
>>> It's a timing belt drive, would that be close to a rack and pinion?
>>>
>>> JT
>>>
>>> On 9/9/2015 11:58 PM, Milosz K. wrote:
>>>> John,
>>>>
>>>> Assuming this is a rack & pinion type configuration:
>>>>
>>>> angular v (w) = linear vel * pulley radius (R)
>>>> angular accel (alpha) = linear accel * R
>>>> Jreflected = (gantry mass + motor mass) * R^2
>>>> inertia total (Jt) = Jmotor + Jreflected
>>>> Torque reflected (Trl) = (Ffriction * R)/efficiency (n)
>>>> Torque peak(Tp) = Trl + ((Jt*alpha)/n)
>>>> P = Tp*w
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> 2015-09-09 14:12 GMT-04:00 John Thornton <[email protected]>:
>>>>
>>>>> Thanks for all the info, now to chew on this a bit.
>>>>>
>>>>> JT
>>>>>
>>>>> On 9/9/2015 5:53 AM, John Thornton wrote:
>>>>>> I need to calculate the power needed to move a gantry. I assume that
>>>>>> mass and velocity and acceleration are the key factors. Knowing those
>>>>>> how do you figure out the watts needed. This is not a milling machine
>> so
>>>>>> no cutting forces need to be taken into effect. I'll be moving the
>>>>>> gantry with timing belts.
>>>>>>
>>>>>> Thanks
>>>>>> JT
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