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How much weight/force to move gear reduction

  1. May 8, 2016 #1
    I need some math help. I am very good with my hands and building mechanical things, but not-so-good with math calculations. ( I got a D in high school physics but can fix damned near anything...) I would appreciate some help figuring this out.

    I have attached a diagram showing what I am trying to calculate, and also some pictures of what I actually built to give the problem some real-life context.

    Basically what I have is a large weight that falls and pulls a cable. The cable is wrapped around a drum which is connected to a shaft. The shaft is supposed to spin a series of gear reductions. ( 9 reductions of 5:1 each. 60 tooth sprockets connected to 12-tooth sprockets with heavy duty roller chain ).

    The idea is to have an initial shaft speed of .00208 RPM input into the gear reduction system that gets ramped up to 4068 RPMs at the output speed of the final gear. That desired input speed is based on the fact that I want the weight to fall 48 inches in 24 hours- which would mean that the 5 1/4" drum would turn 2.9 revolutions in the specified time period. The circumference of the drum is 16.48 " and therefore if the drum turns 3 times ( rounded to 3 from 2.9 ) in 1440 minutes ( 24 hours ) then that equals .00208 RPMs.

    My original thought was to just build the gear reductions and keep adding weight until the darned thing started spinning and the desired output speed was achieved. ( I am so bad at math that I figured it would be easier for me to just experimentally determine the required weight ). However, I seem to have grossly underestimated the required weight / force to move this system.

    As I mentioned, I have 9 sets of gear reductions, and I have added 400 pounds of weight so far and it can only move 3 of them! Some of the parts on this contraption I have built are rated up to 900 pounds, which is great, but other parts are going to have to be rebuilt or reinforced to accommodate additional weight. I have a lot of creaking and groaning and sagging of steel going on... The absolute maximum weight I could use is 1200 pounds but that would require a lot of frame reinforcement and before I start re-building this thing I would like to know EXACTLY how much weight is going to be required to move these gears.

    In the diagram, "A" is the falling weight. "B" is a pulleys. "C" is a cable attached to the weight and wrapped around a cable drum "D". The drum is connected to the first large gear of the reduction system by way of a shaft. Each set of gear reductions goes from a 60 tooth sprocket to a 12 tooth sprocket. The small sprockets are connected by fixed shafts in pairs to large sprockets- which are then connected by roller chain the next small sprocket in the reduction system and so on. Large sprockets are colored blue in the diagram and small sprockets are colored red. The shafts are colored green. I apologize in advance my terrible art skills.

    My question is this- how much weight is required to make the weight fall 48 inches in 24 hours and overcome the mechanical disadvantage of the specified gear reductions and acheive the desired output speed on the last gear? I simply do not know how to determine that without actually building it and adding weight until it moves. I am reticent to proceed because I have underestimated the weight required and the forces involved so far. I am afraid that it is going to require thousands of pounds and i need to know before I build something again that is not up to handling that load. 74337c95d03e5b35c84ed36adb932caa.jpg ab69ca1d4681e8fe504cc04cf5793b2b.jpg 11ee76745bb4b30e6fb61e1829fed436.jpg
     
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  3. May 8, 2016 #2

    billy_joule

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    That is a massive gear ratio. Frictional torque in the later stages is multiplied by the earlier stages so any tiny amount of friction near the output requires massive torque on the input to be overcome. The other problem is that there is a lot of rotating mass which requires alot of torque to accelerate.
    Friction is largely going to determine the rate of descent of the weight and I doubt you'll get repeatable results, have you looked at escapements?
     
  4. May 8, 2016 #3
    Thank you for the reply my freind. Yes, it is a massive ratio, and I really didnt realize the scale of the force required to move it. As I said, 400 pounds was able to move 3 of the reductions, but could definitley not move 5. I was running a test with 600 lbs and 4 reductions when I had to stop because the part that my winch was attached to was about to shear off. I feel like I could have moved the 4th reduction with the 600 lbs. But that's speculation on my part. I know it is going to require a lot more weight to continue to move towards moving the remaining reductions but I have no idea how to calculate how much. Is the force required added or multiplied? For instance, if 400 lbs can move 3 of the reductions, should 1200lbs be able to move 9? If it's multiplied I am screwed because I will need several large elephants. What is an escapement?
     
  5. May 9, 2016 #4

    Bandit127

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    A pendulum clock with a second hand is a mechanism that takes energy from weight that drops over a 1 day period (typically) and turns it in to 1 rpm. While that isn't 4,000 rpm, understanding how one works may give you ideas about how you could improve your drivetrain. It will also explain what an escapement is.
     
  6. May 9, 2016 #5
    Thank you for the reply Bandit. After the original suggestion about an escapement, I did some reading on the subject and now understand what it does. It isnt really applicable to my problem , which is not one of speed regulation or of metering the force out at a given rate. The motive force of this system is gravity, which is constant. The amount of weight will be constant all the way down the length of travel, and the amount of gear reduction will be constant too. I just need to add the correct amount of weight to overcome the inertia presented by the mechanical disadvantage and create the desired input speed ( .0028 RPM ), which will translate into the desired output speed- 4068 RPM. It should all be constant. The question is- how much weight given the known reductions. I can change the weight to any amount that doesnt exceed the structural parameters of the unit. With a little winch and shaft reinforcement, that structural limit will be 800 lbs at least. I may be able to push that to 1200 lbs with even more reinforcement- but won't be able to go higher without a total rebuild and heftier components / steel. At this point, since nobody seems to be able to calculate the correct weight, my plan is to do the easy reinforcement and see if I can get the 4th reduction to move this weekend with 800 lbs or less. I was then planning to incrementally remove weight and experimentally determine the exact weight required to move the 3rd, 2nd, and 1rst reductions as well. With those 4 data points- ie the minimum weight required to move each reduction- perhaps I can extrapolate how much weight will be required to move the 9th reduction by seeing a pattern or factor at play as the number of reductions increases. I just would like an idea of where this is all headed so I have a clear idea of the structural requirements. The good news is that my number 40 roller chain has a tensile strength of 3000 pounds, so that should be able to hold up. My cables and pulleys are rated up to 1500 pounds or higher, but my winch only goes up to 1200 lbs.
     
  7. May 9, 2016 #6

    jack action

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    Maybe I'm wrong, but this is how I would calculate it:

    Let's say that all your stages are identical and produce the same friction torque T just before motion begin. Then:

    A torque T is required to move the first gear set;
    A torque 5*T is required to move the second gear set;
    A torque 5*5*T is required to move the third gear set;
    A torque 5*5*5*T is required to move the fourth gear set;
    A torque 5*5*5*5*T is required to move the fifth gear set;
    A torque 5*5*5*5*5*T is required to move the sixth gear set;
    A torque 5*5*5*5*5*5*T is required to move the seventh gear set;
    A torque 5*5*5*5*5*5*5*T is required to move the eight gear set;
    A torque 5*5*5*5*5*5*5*5*T is required to move the ninth gear set.

    Adding them all up (1 + 5 + 5^2 + 5^3 + 5^4 + 5^5 + 5^6 + 5^7 + 5^8), you need to produce a torque equivalent to 2 441 406*T just to begin motion with nothing connected to any shaft!

    So, if you try to move only one gear set and you need a weight of, say, 0.1 oz (0.00625 lb) to begin motion, you would need 15 259 lb to move all nine gear sets ... with nothing connected to them!

    Another way to look at it is power-wise. A weight of 1200 lb that falls at a rate of 2 in/h produces 0.075 W of power, not counting losses. At 4068 rpm, that much power can only produce an average torque of 0.0000003 lb.in. I don't know what this machine is supposed to do, but that's very, very small.
     
    Last edited: May 10, 2016
  8. May 9, 2016 #7
    Thank you so much Jack Action. You rock! This is exactly what I needed! I had a hard time in high school Physics class- mainly because of the math- and it was such a long time ago. The answer wasn't reallywhat I was hoping for but at least I know the answer now- all I need are 4 or 5 full grown elephants to move the number of reductions that I had wanted to move. ( retired circus animals? ) It's amazing the kinds of forces that can be called into play with just a few gear reductions. Maybe I can drop it to 7 or 8 reductions!
     
  9. May 9, 2016 #8
    Uh.... just so I am clear- if it takes 5 lbs of pressure to move the first reduction, I am going to need 1,953,125 pounds to move the 9th reduction? Almost 2 million pounds? Jesus Tapdancing Christ.... I think 2 million pounds may exceed the structural limitations of this device by just a bit.
     
  10. May 9, 2016 #9
    Anybody out there with better news who would like to dispute the math?
     
  11. May 10, 2016 #10

    Bandit127

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    My point about clocks was more about the way that a high gear ratio is implemented in a way that works. By running dry gears in small (commonly jewel) bearings friction is minimised. Following that logic, if you could reduce the friction you could reduce the starting torque significantly.

    Are you running the chains and sprockets dry? Are the chains plain link or O-ring motorcycle chains? Are the bearings plain or ball/rollers? Are the sprocket pairs perfectly aligned (so that the sprocket teeth have no friction from the chain side plates)?
     
  12. May 10, 2016 #11

    billy_joule

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    Nope.
    If we include more factors it gets even worse than Jack actions analysis.
    As you load up the system the torques required to overcome the various static frictions increases, creating a snowball effect.
    In other words, the act of hanging weights means you start needing even more weights.
    For example, if the radial load on a bearing doubles as you load the system up, the torque required to get that bearing moving doubles, so you need more weights and then the radial load will increase further..

    You could try to reduce friction but I think it'd be futile. If you give us a good description of your goal we may have alternative suggestions.
     
  13. May 10, 2016 #12
    Mr. Bandit- Your point about friction is well taken. The bearings are pillow block style. The chain is well-greased number 40 industrial roller chain such as you might find on a conveyer belt. It has internal bearings and a lubriction reservoir on each one that is full of lithium grease. The sprockets and bearings are attached to solid 1 inch slotted steel shafts with setscrews. I aligned the sprockets with a laser and they are perfect within a milimeter. No side to side friction. The setup is as frictionally efficient as I can make it .
     
  14. May 10, 2016 #13
    Mr. Joule- ugh... you are full of good news. I see your point about the friction not even being a constant for each set of reductions because the act of adding weight itself adds more radial load and therefore more force required to overcome it.

    As far as alternative suggestions, by all means I am open to any. The goal was to use a motive force consisting of nothing but gravity ( in this case, gravity acting on a large, falling, variable weight ) to do useful work at the end of a gear reduction system. And by useful work I mean a shaft spinning at sufficient RPM's and sufficient force to turn a pump, compressor, generator, blower, conveyor or any other device that can use rotational energy as a power source- even a device such as you might hook up to a PTO shaft coming off a tractor or something like that.

    I am a maintenance technician by trade and tinker about in my spare time on any number of ridiculous creations. I just very recently started fooling about with gear reductions- Clearly I underestimated the massive forces that are quickly multiplied in a reduction system. Thank you all for helping me understand how to calculate the required weights to move the remaining reductions.
     
  15. May 10, 2016 #14

    Nidum

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    (1) Large weights which drop large distances are needed to generate useful amount of energy .

    (2) In a multi stage speed increasing gear train the components used can usually be made smaller and lighter in each successive stage .
     
  16. May 10, 2016 #15
    At this point, I need to gather a little more data and figure out exactly what the minimum force/ weight is required to move the 3rd reduction- which was sucessful, but faster than it needed to be as per the diagram above. Once I have that, I can now extrapolate the weights needed for the further reductions. I will also go backwards and see how the data jibes with the second and first reductions. We are dealing with 5:1 reductions, so if I understand correctly,the 2nd reduction minimum inertia-overcoming force/weight should be 1/5th of the 3rd reduction weight, and the 1rst reduction should be about 1/5th of that right? ( setting aside the radial load thing ). The 4th reduction should require about 5X what the 3rd did if I understand correctly. Structurally, that probably limits this particular rack to 4 reduction sets. I doubt I will get to 5. That's assuming the following guestimate weights for the reductions ( haven't confirmed it yet ).

    Reduction 1: 5lbs
    Reduction 2: 25 lbs
    Reduction 3: 125 lbs
    Reduction 4: 625 lbs

    Reduction 5 would be 3125 lbs and there is no damned way. The cable drum capacity is like 800 lbs and the lifting winch is 1200lbs capacity. The tensile strength of the roller chain is in the 3000lb range as I recall.


    whether the speed and force of the output from the 4th reduction is of a useful nature is yet to be determined. I work full time and won't be able to reinforce my winch structure until next weekend so I can't even determine the data for the first 3 reductions right now.

    Maybe gears is the wrong way to go about converting high torque/low RPMs to low torque/high RPMs anyways. What about hydraulics? That's very far from my feild.
     
  17. May 10, 2016 #16
    Well, the weight is limited by the structural strength of the components. Maybe I could focus on the distance that weight falls. Longer travel equals more work I suppose.
     
  18. May 10, 2016 #17

    jack action

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    The first thing you have to focus on is power. The power is constant (no including losses) throughout the mechanical transformations.

    The power P produced by a weight coming down under gravity is:

    P = 0.113 * W * h / t

    Where W is the weight (in lb), h is the height drop (in inches) and t is the time elapsed during the drop (in seconds). The power is in Watts. Divide by 1000 to get it in kilowatts or divide it by 746 to get it in hp.

    Using this equation, if you do not get the power required by the device you want to power with your weight dropping, it will never work, no matter how you play with mechanical advantage.

    And for every pulley/gear set you add, in the best scenario, you will need to multiply that equation by 1.05 (i.e. adding 5%) to compensate for losses.

    This is how I calculated the power of your machine in my previous post. 24 h is 86400 s, so:

    P = 0.113 * 1200 lb * 48 in / 86400 s = 0.075 W
    0.075 W / 1000 = 0.000075 kW
    0.075 W / 746 = 0.0001 hp

    You couldn't even turn a generator to light this 2 W light bulb with that set up!

    if you wanted to light that 2 W bulb with a generator with a 99% efficiency, link to your weight coming down with 3 successive gear sets, the power needed would be:

    2 W / 0.99 * 1.05 * 1.05 * 1.05 = 2.34 W

    Assuming you have a 1200 lb weight dropping from 48 in height, the time required for the drop would be:

    0.113 * 1200 lb * 48 in / 2.34 W = 3274 s

    or just shy of 55 min. Which mean you could light that bulb for 55 minutes before «rewinding» your machine. If you want to use a 100 W bulb to light the room, you will drop that elapsed time to 65 s, i.e. 50 times less.
     
  19. May 12, 2016 #18
    You are correct Mr. Action. Looking at it from the standpoint of power does encompass all of the variables involved. Thank you for showing me how to do the calculations. I can now plug any number of possibilities into your equations and see the results in advance and see what would happen with different weights, times, and reductions. It's hard to believe that 1200 pounds of potential energy can accomplish such damned little useful work. I will have to think some more about how to create useful rotational energy from gravity power. It seems like gear reduction is not the way. What do you think about transmitting much greater force hydraulically? Those sorts of systems contain mad crazy forces internally without the need for support structures and components like sprockets and chains that just cant possibly hold up to the stresses involved with extremes weights and forces like we are talking about with reductions beyond 4 or 5.
     
  20. May 12, 2016 #19

    jack action

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    With hydraulics, you are still only transforming power. You have to think about producing power.
     
  21. May 12, 2016 #20

    Nidum

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