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Calculating spring/motor requirement

  1. Mar 16, 2017 #1
    I'm working on a little ball launcher project where I have to launch a tennis ball 6 to 10 feet. (No more, no less.) I made a spreadsheet to help me quickly try different distance values at different angles. (I know 45 degrees is optimal, but I wanted the flexibility of trying different angles.) The spreadsheet helps me find the required initial velocity. Here are the values I came up with for the required initial velocity:

    oJP7J8H.jpg

    The design I'm using is essentially a crossbow. The ball will be placed in a chair. The chair will be pulled back by a spring that, when released, will accelerate the ball up an incline and lob it the required distance.

    My question is... how do I use what I know (which is the initial velocity required) to calculate the required force constant of a spring? And from there, how do I calculate how far back the spring needs to be pulled? And finally, how do I then determine what kind of motor I need which will have enough torque to be able to pull back the spring?
     
    Last edited: Mar 16, 2017
  2. jcsd
  3. Mar 16, 2017 #2
    Your spring needs to store all of the energy to be imparted to the projectile, plus enough to make up for friction losses in the system. For a simple linear spring, the stored energy is V = (1/2)K*delta^2 where delta is the deflection and K is the spring constant. For a nonlinear spring, find the stored energy by integration.

    There are also matters of how fast you want to release that stored energy and this connects to the cocked displacement you need. It is a systems problem, not necessarily something that has a unique answer.
     
  4. Mar 16, 2017 #3
    I looked up the kinetic energy formula online and found that KE=(1/2)mv^2 where m is the mass of the object and v is its velocity. I know the mass of a tennis ball is 58.5 grams (or 0.0585 kg), and I found the velocity already. So that allows me to calculate the Kinetic Energy (in Joules) pretty easily.

    CAjBdTM.jpg

    So now that I have the energy requirement ... can I use this equation (below) to find the distance that the Spring would need to be compressed? (Let's assume I buy a spring off the shelf which has a given force constant.)

    U=(1/2)kx^2 where x is the distance the spring will be compressed, U is the potential energy (which equals the kinetic energy from the table above), and k is the force constant supplied by the spring manufacturer.

    So to find the distance the spring would need to be compressed, I rearrange the equation for the unknown: x=sqrt(2U/k)

    Is this valid? I'm not terribly familiar with these equations so I'm just basing this on what I'm finding online.
     
  5. Mar 16, 2017 #4
    This is valid, except that you have not accounted for losses and you have assumed a linear spring. The crossbow you spoke of is probably not a linear spring. I'd do some checking on that before going too far.
     
  6. Mar 16, 2017 #5
    Thanks for verifying.

    ...As far as losses go, I know that my calculations are merely ideal. (Frictionless environment and 100% efficient springs.) But since I'm only sending the tennis ball a few feet, I plan on making small adjustments to the hardware as necessary to account for the losses. For example, I'll compress the spring by an extra few centimeters. This project has a low distance requirement, and a relatively large margin for error. By aiming for the center of the target distance (8 feet), I believe I will still be past the minimal distance requirement and won't risk overshooting. But the system will allow for adjustments. For now I'm just trying to understand the equations for the ideal scenario.
     
  7. Mar 16, 2017 #6

    JBA

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    If I remember correctly, in your original thread on this project, you specified that the ball must remain within the 6 to 10 ft area once it lands and that is a key element in determining the amount of horizontal velocity you impart with your launch.
     
  8. Mar 16, 2017 #7
    Correct. It will land on grass, so I am assuming the distance it will roll will be minimal. Targeting the center (8 feet) should minimize the risk of rolling out of bounds one way or the other. That is also why I'm calculating different angles. While 45 degrees is optimal in terms of getting the most distance with the least amount of energy, it may introduce too much horizontal velocity. So that's why I'm going up to 75 degrees in the angle.
     
  9. Mar 16, 2017 #8
    I don't really know what kind of Spring Constant I would be looking for ... so if I assume a compression distance of 4 inches (0.1016 m) (which seems reasonable, I think?), do these values appear valid for the required Spring Constant?

    k=(2U)/(x^2) where x=0.1016m and U= the various KE values in the previous table.

    So for example, 8 feet at an angle of 45 degrees would be =(2*0.5886)/(0.1016^2) is approximately = to 114.04

    kHBZc52.jpg
     
  10. Mar 16, 2017 #9
    As a matter of curiosity, why are you working in this mixed system of units? I see references to feet, inches, meters, gm, kg, etc.

    All of the calculations can be done equally well in any consistent system of units. If you are located in an SI country, then using SI makes sense. If you are located in the USA, then using US Customary Units (inch, lb, sec) makes sense. You will lessen the probability of a mistake if you avoid unnecessary unit conversions.

    PS: In the USC units, with length in inches, the unit of mass is the lb-s^2/in that has no unit name per se, but is a perfectly acceptable unit nevertheless. The pound is a force unit (as long as we avoid the archic unit system using the poundal), not a mass unit.
     
  11. Mar 16, 2017 #10
    The requirements document is in feet and inches. I've converted all length and distance measurements to meters. When I looked up the mass of a tennis ball, google gave it to me in grams. I converted it to kg.

    I am in the USA, but when it comes to physics calculations, all my (limited) experience is in SI. I know the units don't matter as long you stay consistent, but for my own sake, using MKS (Meters, Kilograms, Seconds) is easier.
     
  12. Mar 16, 2017 #11
    You may find SI easier to use (it really is not, but that is beside the point), but when it comes time to buy components off the shelf, do not expect to find them in SI units in the USA. Most of commercial business in the US is still definitely in US Customary units.
     
  13. Mar 16, 2017 #12
    By easier I just mean it's easier to stay consistent with the physics book and the requirement that we use SI in the classroom. As far as the math goes, it makes no difference to me because I let the spreadsheet grind through the repetitive and tedious calculations. I did find it odd that we are required to use SI, but the project was given with US customary units. Typical "Do as I say, not as I do" teaching I suppose.
     
  14. Mar 16, 2017 #13
    I have always found it best to work a problem in the system of units in which it is given. If the given data uses mixed units, then I try for the minimum number of conversions to put everything into a consistent system.

    The one area in which I work where I depart from the above is in electromechanics. The US Customary electromagnetic units are a nightmare, so I always put those problems in SI.

    The fact that the problem was given in US Customary units but you are required to work in SI suggests that the person laying down the requirements has little real world experience.
     
  15. Mar 16, 2017 #14
    I didn't expect this particular point to go past 1 reply :) ... so I should state that in my Physics classes, we always use SI unless a problem explicitly states otherwise. However, this project came from the Engineering class. In my effort to simplify my reply earlier, I intentionally mixed facts from both classrooms together. And I now realize I shouldn't have done that. So to hopefully clarify this completely: As far as I know, there is no strict requirement to convert to SI for this project. But since I otherwise have to do that for my physics class, I was doing that here as well ... since these are all physics equations. So the idiocy falls squarely on my shoulders and no one else's.
     
  16. Mar 16, 2017 #15

    JBA

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    ,What are the units of the spring constant values in your table?
     
  17. Mar 16, 2017 #16

    JBA

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    If I use the conventional US lb/in then the compressed launching force at 45° would be 4 * 114 = 456 lbs which is clearly far in excess of what is required to lob a tennis ball 8 ft.
     
  18. Mar 16, 2017 #17
    joules/m^2
     
  19. Mar 16, 2017 #18

    JBA

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    You need to go back to the energy equivalent equation and calculate the resulting spring rate using the US units that will result in lb/in rate values if you want to be able to check available springs from US suppliers.
     
  20. Mar 16, 2017 #19
    There is certainly no need for any sort of apology. I was simply suggesting what I have concluded after a long career (almost 50 years) doing this sort of stuff on a daily basis.

    You said earlier that M = 58.5 gm, which translates to
    W = 58.5/1000*0.4535924 [=] gm*(lb/gm)
    = 2,653516*10^(-2) lb
    and
    M = W/g
    = 2.653516*10^(-2)/(12*32.174)
    = 6.872826*10^(-5) lb-s^2/in (if I have done all the arithmetic correctly!).
    This is the value to be used in any kinetic energy or F = M*a calculations if US Customary units are used.
     
  21. Mar 16, 2017 #20
    I made a copy of the sheet I was working on and tried to convert it all to feet and pounds. (Condensed the format a bit for simplicity.) The Initial Velocity in ft/s seems right to me. (And a quick google conversion agrees.) What about the Kinetic Energy and Spring Constant tables?

    elqMVuw.jpg
     
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