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Spring constant to launch ball 70 mph?

  1. Jan 9, 2009 #1
    Could one of you mechanical geniuses please help out a dumb ol' elec engr? I've searched the boards for a few hrs now, but can't seem to get confidence in my answer to what should be a simple problem.

    I'm trying to build a simple a simple tennis ball launcher to fire a ball out at 70mph -- assumptions below. I've found v= sqrt(k/m)*l as the equation I apparently need to use. So, I can get a "number", but I am not clear on how the units all work out as I am not seeing a time unit on the right side of the equation.

    I would REALLY appreciate someone knocking this out for me so that I can at least get in the right hemisphere for starting to experiment with diff spring sizes.

    So, here are the relevant parameters & constraints:

    1. Launch mechanism is a simple, linear compression spring loaded, then abruptly released to launch a tennis ball out of the barrel of the apparatus
    2. Tennis ball mass = 5 oz.
    3. Desired velocity of tennis ball upon separation from fully recovered spring is 70 mph
    4. Spring to be compressed exactly 5.0 inches
    5. Assume no frictional or angular losses
    6. Assume no elasticity/rebound effect from tennis ball construction.

    Question: What spring constant, K, is required to accomplish the 70 mph launch?
  2. jcsd
  3. Jan 9, 2009 #2
    F=ma and F=kx. Set equal to each other, kx=ma, and use v^2=2ax (v initial = 0). Solve for a from the first relation, a = kx/m and substitute into the velocity equation and solve for k.

    k = v^2*m/(2*x^2) where x is the distance the spring is compressed.
  4. Jan 9, 2009 #3
    Thanks, Chris.

    I'm still confused on units, however. With a couple of conversions to get matching units, my parameters are:

    v desired 1232 in/s
    m 0.3125 lbs
    x 5 in

    Based on the equation k=v^2*m/(2*x^2), I get:

    k = 9486.4 ??Units?

    Or, do I need to put something else in diff units prior to calc?
  5. Jan 9, 2009 #4


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    k has the units of spring stiffness so from F=kx, k must be force/length ie N/M in SI.

    You could also use conservation of energy
    Spring energy = 1/2 kx2
    Kinetic energy = 1/2 mv2
  6. Jan 9, 2009 #5
    mgb -- yes, I understand that k is a force/length unit. The problem I'm having is that I believe the equation above, when used with the units (lbs, inches and s) I've shown, reduces to something like lbs/sec^2.

    What am I missing?
  7. Jan 9, 2009 #6
    mgb -- using the conservation of energy equations I get double (18972.8) the number of previous calc. What gives?
  8. Jan 9, 2009 #7
    Thats not quite right Chrisk, since acceleration is not constant...
    Here is how should be (though it is only a factor of two that is the difference), I will be more explicit so you can understand it.

    Let the origin of our frame be at the point where the ball leaves the tube. At this point we want the velocity of the ball to be the greatest. (We neglect aerodynamic, and frictious effects..)
    Now the equation of motion of the ball will simply be that of a harmonic oscillator. That is:


    Where k is the spring constant we are looking for, and m is the mass of the ball.
    Rearranging we get:

    Now lets rewrite the second derivative as follows:


    But the first time derivative of the displacement is velocity hence: [tex]v=\frac{dx}{dt}[/tex]

    So our differential equation is:
    [tex]\frac12 \frac{dv^2}{dx}=-\frac{k}{m}x[/tex]

    This is a simple Differential equation for the v(x) velocity-place function. As we see it is separable.
    If the spring is compressed by [tex]L[/tex] (in your case 5in), then the velocity will be zero at -L that is v(x=-L)=0. And you want it to be a fixed value v at the end of the tube so the integration is:

    [tex]\int_0^{v}dv'^2=-\frac{2k}{m}\int_{-L}^0 x ; dx[/tex]

    performing the integration:


    So the spring constant you need, if you want to compress it by L (5 in), and want the tennis balls velocity to be v (70mph), and the mass of the tennis ball be m:

    Last edited: Jan 9, 2009
  9. Jan 9, 2009 #8


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    In the force version have you accounted for the fact that when the ball is released the spring has no compression (and so no force) the full spring force is only when the spring is fully compressed. Remember F=kx is only for the x at that point, what is the average force?
  10. Jan 9, 2009 #9
    Thaakisfox -- wow, thanks for spelling it out for me. Using your resulting equation, I got the same # as I did with mgb's suggested cons of energy approach.


    v desired 1232 in/s 70 mph
    m 0.3125 lbs 5 oz
    x 5 in

    k= 18972.8

    Not to be a complete moron, but I still have no idea what units the answer represents. Anyone??
  11. Jan 9, 2009 #10


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    It will be force/distance so lbf/in.

    Energy conservation is always useful when things are changing since you don't have to worry about the intergrals - just energy at start = energy at end.
  12. Jan 9, 2009 #11
    Wow, so is this really saying I'd need almost 19K lb/in for the spring constant?
  13. Jan 9, 2009 #12
    Thaakisfox did a nice derivation. I apologize for assuming the acceleration is constant. Now, for the units. The issue is mass in English units is the slug and force is slug-ft/sec^2 which is the unit of pounds. Spring constant is usually expressed in force/length e.g. in MKS units Newton/meter. So, convert the mass to slugs, and the length to feet. Since W = mg, m = W/g where g = 32 ft/s/s and W is in the units of pounds. The English units for mass-length-time is slug-foot-second. After doing the conversions your units for k will be slug-ft^2/(sec^2-ft^2). Noting that a pound is a slug-ft/sec^2, K will be in pounds/ft.
  14. Jan 9, 2009 #13


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    70 mph is quite fast even for a tennis ball and 5 inches isn't much of a stretch.
    But I think thats rather a lot
    Last edited: Jan 9, 2009
  15. Jan 9, 2009 #14
    Amazing, this fuss about the units... :D:D
    Why cant it be just SI everywhere in the world, why does it have to different everywhere.. lol
  16. Jan 9, 2009 #15
    OK, using the correct conversion and the correct expression for k we have:

    m = .3125/32 = 0.0097 slugs

    v = 70 mph = 102.67 ft/s

    l = 5 in. = .4167 ft

    k = (0.0097)(102.67)^2/(.4167)^2 = 589 lb/ft = 49 lb/in.
  17. Jan 9, 2009 #16


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    Or in SI
    x = 5inch = 0.127m
    m = 5 oz = 0.141 kg
    v = 70mph = 31.3m/s

    k = mv^2/x^2 = 0.141 * 31.3^2/(0.127^2) = 8500N/m = 49.1 lbf/in
  18. Jan 9, 2009 #17
    Thanks a ton for all the help, guys. I did acutally convert to the SI values as mgb and miraculously (sigh, so simple) got the same result.

    Now the part that is freaking me out is the additional force it's showing I'm gonna need if my spring retractor/plunger take that mass up to around 20 oz.. Under that scenario I calc needing k = 196 lb/in. Compressing that 5 inches = YIKES!

    My hope is that real testing will reveal that the rebound of the compressed ball will take these forces down an order of magnitude. Any thoughts on that liklihood??

    Thanks again for all your help.
  19. Jan 14, 2009 #18
    You say that the spring is to be compressed 5 inches. Do you mean that the spring is to be completely free when the tennis ball is launched, or do you mean that the stroke of the launcher is only 5 inches? If the spring can remain partially compressed when the ball is launched, this is a different problem.

    Second problem I see with the work presented above is the units used for mass. It appears that you have expressed the mass in pounds, but the proper mass units for this calculation would be in units of lb-s^2/in; pounds represent weight.
  20. Jan 14, 2009 #19


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    Only Myanmar, Liberia, and the USA, along with a few hold-outs like the UK, use non-metric units. Don't hold your breath waiting for the USA to convert any time soon. The movement to convert the USA to metric started some 40 years ago. It made some minor progress in the early years, but for the most part, it is now all but dead, even in the engineering fields (especially civil/structural, where pounds, feet, and inches will be around for at least another 40 years.) The feds tried to set the example about 10-15 years ago by producing highway construction plans in metric units.....there were so many costly errors made, that the plans are now back now to being prepared in the good old fashioned USA units.
  21. Jan 14, 2009 #20
    The reason for the fuss about units is that you have to get them correct if you want to get a useful result. Lots of the work posted on PF seems to involve messed up units, particularly when people are using US Customary units and they need to express mass. This is an old problem, and it does not appear to be being handled any better, perhaps not as well, as it was 40 to 50 years ago.

    As to why the metric conversion of the US seems to have stalled, there are lots of possible answers. Ultimately I think it was never presented in such a way as to be profitable for business to go to SI, so they did not make the effort. Business is in business to make money; that is always the bottom line. If a particular idea does not contribute to making more money, it will not be adopted. Many parts of US industry have converted to SI, but many parts have not, which makes for a real mixed bag. Places like Fastenal, the nut and bolt suppliers, carry both USC and SI sizes for everything.
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