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Thread: Motor help

  1. #1
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    Confused Motor help

    May someone teach me how to choose a dc motor?

    If I want to lift a person up using scissor lift drive with power screw. As common the torque required will be highest at the lowest position before lifting, Is it I should use the torque in this position to calculate the power required.

    One more is how to determine the speed required? As i know when lifting, torque will reduce is it the rpm will increase for dc motor? So torque and speed are variable along lifting? So how to determine the torque and speed required to find the power ? How to base on torque and speed to choose a motor in manufacturer catalogue? Any extra infomation needed?

    Finally i a gearbox is used for the speed reduction after choosing a motor in here.

    Thanks for any one to share his idea. Due to I am not very familiar in electrical motor, so hopefully can get a share from this forum.

  2. #2
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    Quote Originally Posted by Solution View Post
    Is it I should use the torque in this position to calculate the power required.

    One more is how to determine the speed required?
    The lifting speed and load will be defined by you. How fast do you want to lift the maximum load at motor maximum speed? Work back from there using the pitch of the lifting lead screw and any gearbox or drive belt ratios. Select a motor with that speed range.

    For sizing the motor, calculate the max Horse Power required with the above ratios in mind.

  3. #3
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    motor

    thanks,
    From the static analysis, the lowest position scissor lift with horizontal force around 2142 N thus the torque in power screw in this maximum torque is 9.74Nm. If the vertical lifting speed i assume to be 1.4cm/s which the total distance lift is 14cm while i set the time to 10s to complete.

    How about the horizontal speed for the power screw determination? I know that 1 rev will equal to the lead distance of screw, but the speed in vertical 1.4cm/s how to determine horizontal from vertical speed desired? during lifting, the load decrease so the torque too,so the rpm speed will increase?

    A DC motor is constant power or constant torque or constant speed supply to the system?

    So selecting a dc motor, we find the motor spec according to the speed calculated? any other factor to choosing a motor?
    I heard that the torque vs rpm curve for a motor can find a best efficiency for it but i have no idea how?
    please help.thanks
    Last edited by Solution; 03-09-2011 at 12:15 PM.

  4. #4
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    I would get the highest mechanical advantage and the lowest, take the average from which

    ME=(MEmax+MEmin)2
    MEmax=dy/pitch=dy/p at lowest position
    Let
    dy vertical motion for 1 revolution of jackscrew
    T*2pi=F*p=W*dy energy equation

    F= thrust on jackscrew
    T= torque
    pi=3.14159
    W= vert load


    ME=F/W=dy/p

    Now you have to allow for static friction

    We increase T and F by 30 % for this; then

    T=1.3*W*p*ME/pi

    Next you need the speed of the motor in RPM
    (dy/dt)/(p*ME)*60

    To get power:

    From the 1st eq, using 1.3W instead of W to account for friction (also acceleration to start but not included for DC motor due to high starting torque)
    T*2pi=1.3*W*dy
    Divide by time dt to get the power, where dt is the time to lift dy= time for 1 rev of jackshaft.
    T*2pi/dt=1.3Wdy/dt
    The RHS is the power requirement s and you know all the values
    dy/dt=.1.4cm/sec
    W =vertical load on lift, Newtions
    You can do the conversion for HP

  5. #5
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    Thanks, but how to get mechnaical advantage for min,MEmin=?

    MEmax=dy/pitch, dy is the vertical motion for 1 revolution of jackscrew, dy is get by the proportionality of,
    total vartical displacement/total horizontal displacement = dy / pitch of screw?

    dy/pitch , the pitch is it the pitch of the screw?
    F= thrust on jackscrew in my case is it 2142N? but i have change it to torque need to rise for power screw ,equation
    Tr= F*d/2[(l+pi*f*d*sec (alpha))/(pi*d-f*l*sec (alpha))] where f=friction, l=lead of thread screw, alpha= helix angle screw, d= mean diameter.
    but why from your energy equation T*2pi=F*p the thrust force change to torque is different to mine?

    Sorry for asking a lot of question. thanks for your reply

  6. #6
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    answer

    "Thanks, but how to get mechnaical advantage for min,MEmin=?"

    Same way you got MEmax, from the linkage analysis. BTW, the highest mechanical advantage is at the lowest position.

    "MEmax=dy/pitch, dy is the vertical motion for 1 revolution of jackscrew, dy is get by the proportionality of,
    total vartical displacement/total horizontal displacement = dy / pitch of screw?"

    I have used pitch for your lead ,;; l assumed single thread. Your total vertical displacement /total horizontal displacement is the average dy/p or dy/l

    "F= thrust on jackscrew in my case is it 2142N? but i have change it to torque need to rise for power screw ,equation
    Tr= F*d/2[(l+pi*f*d*sec (alpha))/(pi*d-f*l*sec (alpha))] where f=friction, l=lead of thread screw, alpha= helix angle screw, d= mean diameter.
    but why from your energy equation T*2pi=F*p the thrust force change to torque is different to mine?"

    Your T is not my T. My T is torque= your T*r
    It is different since mine does not include friction. If you make f= 0, and Tr=T, you will get my ideal energy equation. Nothing wrong with your equation but it applies only to the screw and misses the bearing friction. I have lumped all the friction together and made a broad assumption that the effect o friction and safety factor is equivalent to increasing the payload by 30%..

    BTW, it looks like you are pulling equations you don't fully understand. You either need a consultant to help you through this or get a better understanding from engineering textbooks on the subject. I don't think this forum can do all of your engineering, but if you can get a better grip on the subject we can help.

  7. #7
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    Thanks a lot...now i understand...sorry for my ignorance,can you teach me how to calculate the mechanical advantage..? because my scissor lift is assymettric as it lifting, the arm will not directly move upward vertical..it will move in tilt..

    Like the files show in http://www.fileupyours.com/view/303536/lift.bmp
    Can you show me how to get the dy? if my lead is 1.75mm. or give me example for calculationof scissor mechanical advantage? Please...thnaks a lot

  8. #8
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    Where is the tilt?
    ME=tan (beta)
    beta= angle of transverse link to horizontal
    L/dy=ME
    dy=L/ME
    BTW, I had this inverted in my original post
    and you were right ;for this linkage, the lowest position has the smaller ME and the ratio of high to low is about 4, so in this case the torque at the low position is highest .and would important for the motor design.

    I might add your design is more complicated because of the 4:1 range of mechanical advantages. Servo speed controlled PM motor + gearbox is needed. and the motor power is minimal since you basically have to lift the mechanism plus the person a small height in 10 seconds plus friction.

  9. #9
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    Thanks again for your reply....and your knowledge...
    Sorry for that, actually for tilt is inclined....because i consider one of the linkage as calculation first, at the bottom end of linkage when it move horizontal with a lead screw distance the top point will move upward not vertically..but inclined with small angel. which can be seen comparison between initia condition and end condition.of the point. But i think it is not the point, we can treat it lift vertically.

    Because it is lift vertically for 1:1 range of mechanical advantages scissor lift.
    Is it ME=tan (beta) can use in this lifting type too for my 4:1 range of mechanical advantages?
    torque at the low position is highest .and would important for the motor design so we only consider this torque and
    speed we take the average of ME=(MEmax+MEmin)/2 or [dy/dt (at lowest position) + dy/dt(at uppest position)]/2 ,what should dt for 1 rev?

    After all calculation, the motor choosing base on torque and speed calculator? or multiply up to get power then only choose from catalogue spec?
    thanks

  10. #10
    Technical Fellow jboggs's Avatar
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    Check the current issue of Machine Design magazine or this link:
    http://machinedesign.com/article/mot...made-easy-0302

  11. #11
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    My final answer on this

    1) Get the maximum torque required at the lowest position. I do not think you need a gearbox .

    2) then get the energy it takes to lift the mass and frame (potential energy); divide that by the total lift time to get the average motor power
    .
    3) Look for a motor that is sized about 50-75 % more than you get in (2) (to account for friction and safety factor) and has a stall torque that amply exceeds the value you got in (1)

    4) Finally make sure the no load speed exceeds the highest required speed by a good margin to handle the lift speed near the highest position

    Now that only gets you a candidate motor.

    5) You next have to hire a competent servo consultant who will design a speed and position control system that meets your time requirements of almost constant speed smoothly coming to rest at the final position without impact .
    The consultant should also be able to help you choose the motor.

    Good luck.

  12. #12
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    Thanks.

    zake, jboggs and PinkertonD, thanks alot...appreciated for your guides and helpful advices and also steps to choosing a motor. Very thanks
    Last edited by Solution; 03-12-2011 at 10:57 PM.

  13. #13
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    I know the how to get the ME and know the reason to get it...but now i doubt is the speed in highest position...because for dy/dt/(MEmin*lead)*60 will get highest speed....May i know how to calculate the highest speed in highest position in my case that cannot exceed max speed of motor..? thanks..

    The torque calculated from W*dy=T*2pi, is very highest compare to what i calculated using screw torque rise equation..why?
    W*dy i know is mgh or mass *gravity * total height lift =potential energy.
    Last edited by Solution; 03-16-2011 at 08:52 AM.

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