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Need enginering help - lifting 1000 lbs 16 feet

  1. Nov 17, 2013 #1
    Need enginering help -- lifting 1000 lbs 16 feet

    Ladies & Gents,

    I am trying to engineer a lifting device. I can build the frame and all that but getting the motor and gearbox sized right is my problem. I am not a math wizz. Some algebra in middle school many many years ago .

    The load I want to lift is 1000 lbs. The vertical travel distance will vary from 8' to 16'. The travel speed need not be very fast, 12 inches a min is plenty because there may be binding issues during said travel. I had figured to use a 3" sprocket and chain on the gearbox to do the lifting with. What I don't know is how to translate that 1000 lbs into a motor size. Keeping the motor/gearbox assembly light is one of the desired design features as it will need to be moved by hand repeatedly.

    I have looked on google and found lots of info but translating that into actual useful numbers is beyond me.

    Can anyone help me?

    Thanks
     
  2. jcsd
  3. Nov 17, 2013 #2
    I am sure my math has problems but this is what I have at this point.

    Sprocket radius
    0.125 feet
    Lift Weight
    1000
    Ft Lb torque
    .125*1000 = 125
    Sprocket Circumference
    2*PI*.125 = 0.7854 feet
    Lift Height
    16
    Lift Time (min)
    15
    Output RPM
    15/(16/.7854) = 0.7363
    Input RPM
    1750
    Gear Ratio
    1750/.7363 = 2376.714
    Horse Power
    .7854*.7363/5250 = 0.01753

    I have this set up in a spreadsheet but when I change the lift time from 15 to 30 min my ratio goes in the wrong direction. I have been using the formulas here. http://machinedesign.com/technologies/gearmotor-sizing-guide

    if my ratio is correct I will need to further reduce the ratio by chain & sprocket, not really a problem.

    Thanks
     
  4. Nov 17, 2013 #3

    SteamKing

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    It's not clear from your post if you are using any tackle (pulleys and sheaves) in your lifting device. There's more to engineering than throwing a bunch of numbers on a sheet of paper.
     
  5. Nov 17, 2013 #4
    Since I didn't mention them, no there are none, besides I have never seen them work well with roller chain ;).

    Having though about what I want to do and the best way to do it I chose roller chain over wire rope. I considered using screw jacks but getting an adjustable screw length for the different distances I need to move the load is problematic with out having separate screws for each length and needing as many as 18 of each that gets expensive. Wire rope has its own problems, fraying, kinking and without using block & tackle heavy and hard to manage.

    With #60 roller chain I can run one chain from the lifting point to the lifted point and pull. Very simple. Also rotational/twisting of the lifted load is not an issue although binding may be. If I need more lifting length I can add chain easily with a master link. When the lift is done the chain can be rolled up and stored in a 5gal bucket. No sheaves (shivs) for the wire rope to come off and be damaged by and no screw to get damaged during storage and / or assembly.

    Still you didn't mention if there was a problem or lack of problem with my math.

    Thanks

    I actually have a crude dxf drawing drawing of the basic design but I hesitate to upload it because it contains sensitive information.
     
  6. Nov 18, 2013 #5

    SteamKing

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    Well, in that case, I understand.

    Still, trying to hoist something slowly when your motor is turning at 1750 RPM is tricky, if you don't use some kind of reduction gear to slow things down.
     
  7. Nov 18, 2013 #6
    From my earlier post
    As far as that goes I do not have to use a 1750 motor. They are common and it theory, cheaper. The mechanical part is NOT the problem. The problem is getting the math on the gear motor right so that I get units that are big enough but not excessively big and heavy as they have to be carried by hand up and down a ladder in soft sand.

    Understanding how the math is suppose to work is my problem. And still I am only getting responses about the mechanics.


    FYI: Moderators; this site seams to be trying to load something that it is unable to download, even after setting all night it still does not complete. At least in Opera on Debian Linux. Also I am behind a firewall that drops all data to and from know hacked sites. See http://www.spamhaus.org
     
  8. Nov 18, 2013 #7

    Baluncore

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  9. Nov 18, 2013 #8
    Thanks Baluncore

    My grandpa actually had one of those now that I think about it. But that wont work very well for me in this app because the drive unit actually needs to climb the chain, not the other way around. Quality #60 roller chain has a 7000 lb tinsel strength. Plenty strong enough to lift 1000 lbs me's a thinkin. Also this is not a continuous duty device, if it gets used once a month over the coarse of a year it will have been a real good year.

    Still no one has mentioned why the math posted above is not working as I think it should.

    On another note; when a supplier of gear boxes lists a horsepower rating is that rating normally for the input or output?
     
  10. Nov 18, 2013 #9

    jack action

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    The work you want to do is pulling a force (F) at a desired speed (V). The power (P) you need is:

    P = F*v (in metric units)

    1 lb is 4.448 N and 1 m/s is 2362 in/min, so: 1000 lb is 4448 N and 12 in/min is 0.00508 m/s, hence:

    P = 4448 N * 0.00508 m/s = 22.6 W or 0.03029 hp (1 hp = 746 W)

    You will have to add some losses to that power (like friction), but this is you basic number to size your motor.

    A couple of mistakes in this. Let me redo your work:

    Sprocket radius
    0.125 feet
    Lift Weight
    1000
    Ft Lb torque
    .125*1000 = 125
    Sprocket Circumference
    2*PI*.125 = 0.7854 feet
    Lift Height
    16
    Lift Time (min)
    15
    Output RPM
    (16/.7854)/15 = 1.3581
    Input RPM
    1750
    Gear Ratio
    1750/1.3581 = 1288.565
    Horse Power
    125*1.3581/5252 = 0.03232

    As you can see, it gives the same number as previously calculated (It's a little higher because 16 ft in 15 min is a speed of 12,8 in/min instead of 12 in/min).

    The power rating should represents the input power, as it is always the largest of the two.

    Although, in theory, the input power is equal to the output power.

    That is the beauty of the notion of power: Power stays constant throughout any process (less some losses like bearing friction). What goes in must comes out. So the input power is the same as the output power, only the torque and rpm are different (Tin * RPMin = Tout * RPMout).
     
  11. Nov 18, 2013 #10
    Thanks jack action,

    You have been much help. I got the formulas fixed in my spreadsheet and now it calculates motor sizes as it should.

    If I am understanding you correctly this gear box would be suitable for what I am trying to do.
    http://www.surpluscenter.com/Power-...1-RA-GEAR-REDUCER-0-6-HP-IEC-71B14-LEESON.axd
    SPECIFICATIONS
    Ratio 60:1
    Power 0.6 HP max.
    Input Speed 1750 max.
    Output Speed 29.16 RPM max.
    Torque 445 in.-lbs. max.
    Motor Mount IEC 71B14
    Input Shaft 14mm keyed hollow input bore
    OutputShaft 1-1/8" diam. x 3-3/16" thru length hollow output bore
    Mount Four tapped mounting holes on each output side
    Size 5-1/2" x 6-1/2" x 4-3/16"
    Shpg. 9 lbs.

    All I would need to find is suitable motors to go with them and gear it down a little more with an extra set of step sprockets.

    Thanks
     
  12. Nov 18, 2013 #11
    amp = hp . look into electric motors gear diver
     
  13. Nov 18, 2013 #12

    Baluncore

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    I agree with those numbers.

    The Lift.
    mass of 1000 lbs = 453.59237 kg
    height 16 feet = 4.8768 metres
    acceleration of gravity = 9.80665 m/s/s
    force = 453.59237 * 9.80665 = 4448.22 newton
    total energy needed is = 4448.22 * 4.8768 = 21693. joule

    Motor Power.
    1 HP = 745.7 watt
    15 minutes = 15 * 60 = 900 second
    Power for 15 minutes = 21693. / 900 = 24.1 watts = 0.032 HP
    If lifted in 30 minutes = 12.05 watt = 0.0162 HP
    If lifted in 5 minutes = 72.3 watt = 0.097 HP

    Gearing.
    sprocket radius is 0.125 feet
    Lift is 16 feet. Motor is 1750 RPM
    Total turns = 16 / (2*Pi*0.125) = 20.37 turns
    20.37 turns in 15 minutes is 1.358 RPM
    Gearbox ratio needed is 1750 / 1.358 = 1288.66

    By looping the chain over a top sprocket you halve the chain strength requirement and halve the gearbox ratio required to 644.33

    By using a differential pair of sprockets say 32 and 33 tooth, you get another factor of 32 reduction to a gearbox ratio of 19.83
     
  14. Nov 18, 2013 #13
    Go to almost any industrial gearmotor supplier site. Most offer a great catalog with Application Engineering section or a standalone AE handbook. Some even have online calculators that will select your gearmotor for you. SEW Eurodrive always had the most rigorous engineering documents in my experience.

    Determine your required speed profile. You size your gearmotor for peak torque which will be torque required to accelerate the load from zero speed to max speed in the required time + torque to lift 1000 lbf against gravity + torque to overcome friction + etc etc etc. "Load" will be all the rotational & translational inertias that you must accelerate.

    Gearmotors are de-rated based on how many starts per hour. More starts, more heat, bigger motor required.

    You'll need a suitable brake to hold the load from falling.

    Unless you have all of the mechanical elements already figured out, sizing a gearmotor is usually an iterative process. It requires a few tries to converge on a good answer.

    Torque is cheap. Buy plenty of it.
     
  15. Nov 19, 2013 #14
    A preliminary way to determine what motor size you're going to need is to just solve for the energy requirements. You want to lift 1000lbs 16ft, so the energy required is 16,000ft-lbs. You also want to lift at 12"/minute (1 ft/min), which means it will take 16 minutes or 960 seconds to lift. So we solve for power: 16000ft-lbs / 960 seconds = 16.67 ft-lbs/sec. Remembering that one horsepower is 550 ft-lbs/sec, you need a motor that can deliver 16.67/550 HP or 0.03 HP. But you will probably need to double that in order to compensate for the losses in the transmission and lift mechanics.
     
  16. Nov 19, 2013 #15
    DANGER! I want to add one important point here and that is a chain and sprocket transmission used with an electric motor can be extremely dangerous in a lifting application. If the motor should lose power, there's nothing to stop it from free-wheeling allowing your load to crash to the ground. A worm gear drive is much safer (although not as efficient). Even with a worm drive, a brake should also be incorporated.
     
  17. Nov 19, 2013 #16
    Thanks to all.

    So 1 tooth difference in sprocket size divides by 32? Some how that does not seam right to me. Please explain.

    Another concern I have is the shaft that will hold the main lifting sprocket. The bearing centers will be 7.5 inches apart. There is not much I can do practically to make that narrower. How big a shaft do I need? Sure I can just through an 1.25" shaft in there but such a big shaft limits how small the lifting sprocket can be. If I could use a .75" shaft I could use a smaller lifting sprocket and further reduce my required ratio.

    My dad is missing fingers on his right hand because an 8N Ford tractor he was working on fell, even though he thought he had it properly blocked up. Needless to say I grew up learning a very healthy respect for things that can fall.

    Just the other week we were fixing a freight lift that the slack-chain safety stops had not worked correctly on and it broke the chains. One of the electrician's helpers was standing down below and claimed he could jump out of the way if we lost the chain again while we were putting it back on. I made him get out of the fall radius anyway. Sure enough not 5 min later the other chain slipped and went crashing down right where he had been standing. Point, YOU CAN NOT MOVE FAST ENOUGH WHEN SH** HAPPENS.

    Yes I am planning to put a spring loaded mechanical pinch brake/stop in the system. Even if it did not hurt somebody I still can't afford $15000.00 to repair/replace the damaged material such a disaster would cause.
     
  18. Nov 19, 2013 #17

    Baluncore

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    Maybe you have not understood differential gears with an endless chain in two loops. The two sprockets are locked together on the same shaft so one revolution of the sprockets takes in 31 links but releases only 30 from the supporting loop. That shortens the loop of chain by one link per rotation of the sprockets.

    You have the two falls of the supporting chain loop, not just only the one supporting the load. The chain shortening is in a loop of chain so you keep the factor of two reduction by having an idler. It also halves the weight of the chain required.

    If you put the differential pair at the top and the idler at the load then you can drive the idler to lift the load if you do not want to drive the differential pair above.

    Another advantage is that you can use safer lifting chain that is much stronger and cheaper. Link chain is less likely to run without motor power than roller chain. Because it is easier to lubricate and is less liable to damage than roller chain it is not so certain to result in a catastrophic failure later. Remember, a chain is as strong as it's weakest link.
     
    Last edited: Nov 19, 2013
  19. Nov 19, 2013 #18
    Thanks

    Yes I do, but I did not realize that that is what you were talking about. But thanks for the through explanation of how they work.

    The chain may be cheaper but I have yet to find anyplace to buy the sprockets. You have a US supplier?

    The Differential sprockets are a good idea but they have a problem that I don't see an easy way around. Just like the screw jack method needing a different screw for each different lifting height the differential sprockets would need a different length chain for each different lifting height. Unless you know a way to magically grow and shrink a welded link chain. : ) As I understand it the idler sprocket used to drive the differential sprockets above must remain in a fixed location to keep the slack out of the chain. On some other method of keeping the slack under control must be devised. Simple is better. Less things to go wrong. KISS

    Take a look at this video that shows the goal but not the method
    http://www.google.com/url?sa=t&rct=...pIp2N3xm9TfRTf0hQnswiNg&bvm=bv.56643336,d.b2I

    His method has a single point of failure (the winch on the dozer) with catastrophic results. My plan is to put a separate lifting device at each post with them all controlled from a single location. That way if any one fails the the whole thing does not come down. Also note that his barn design is OK for the southern US but will not work in the north where we have lots of snow.

    Obviously putting anything very heavy and bulky on top of the posts is problematic and dangerous from a ladder. Often it is not possible to get a man lift in place for putting the units on the post, like the site being off grade or to sandy for them to get around, trees in the way, the list is long. Hence the lifting unit needs to actually be under the header where it can be installed from the ground. Removal after the lift can be achieved with a simple rope thrown over the truss and lower it to the ground.

    Thanks
    Randy
     
  20. Nov 19, 2013 #19
    Make a pole climber using a rack and pinion drive. Attach the rack to the side of the pole and have the pinion drive attached below the rafters. The pinion drive walks up the rack lifting the roof.
     
  21. Nov 20, 2013 #20

    Baluncore

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    With the differential pair at the top and the idler supporting the load there is one loop under tension and another loop hanging freely below. The length of the endless chain remains fixed. To lift 6 metres or less requires 12 metres of chain closed into a loop.

    I am not in the US so I don't have a US chain sprocket supplier there. You can do a google image search for "round link chain sprockets". Take a look at these.
    http://www.pewagchain.com/Products/...ettenrader-aus-Guss/IR-Zahnkettenrad-(2).aspx
    I have made chain sprockets by cutting a deep groove in the periphery of a wheel for the chain edges and then milling recesses for the links with a circular cutter.
     
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