I am trying to decide on a constant force spring to be used in a projectile launcher. The launch velocity will need to be approximately 15 m/s, which, with a barrel travel distance of 1 ft, equates to a constant force of about 20.6 Newtons (about 4.6 lbs) to reach that velocity if my calculations are correct. So my question is, where do I go from here? When I look up constant force springs, it tells me what their maximum load is, so I assume that is what the constant force will be if the spring is stretched to it's maximum length. So is it really as simple as choosing a spring with a maximum load equal to my desired launch force, or is there something more to consider? And how would I go about choosing the length and width of the constant force spring, as I'm sure it needs to be a certain distance longer than what will actually be used?


2 Answers 2


Provided your calculations are correct (you are not mentioning the mass that you will be launching) then you are right you need nothing more. However, I am not sure if its easy to find a constant force spring with a travel of 1 foot.

If you are using a common spring (that follows Hooke's law ) then you can easily estimate the spring's constant using the energy equilibrium

$$ T_{0} + V_{g,0} +V_{s,0} = T_{1}+ V_{s,1}+V_{g,1} $$

Since you are launching at 45 degrees and the extension of the spring is 1 foot then

  • $T_0 =0$ is the kinetic energy initially (V=0)
  • $T_1 = \frac{1}{2} m v_{1}^2$ is the kinetic energy when the spring has $\delta x =0$ , ($v_1=15[m/s]$
  • $V_{g,0} =- mgh$ is the potential energy due to gravity initially ($h = \frac{2}{\sqrt{2}} \cdot 1 [ft]$)
  • $V_{g,1} =0$ is the potential gravitational energy when the spring has $\delta x =0$ (h=0)
  • $V_{s,0} =\frac{1}{2} k \Delta x^2$ is the spring energy initially ($\Delta x= 1ft=0.3m$)
  • $V_{s,1} =0$ is the spring energy when the spring has released ($\Delta x =0$ , h=0)


$$0 + (- mgh) +\frac{1}{2} k \Delta x^2 = \frac{1}{2} m v_{1}^2+ 0+0$$

$$ k = \frac{2m}{\Delta x^2}\left(\frac{1}{2} v_{1}^2+ gh\right)$$ $$ k = \frac{2m}{\Delta x^2}\left(\frac{1}{2} v_{1}^2+ g\cdot \frac{2}{\sqrt{2}} \cdot \Delta x\right)$$

  • $\begingroup$ I failed to mention that this is a tennis ball (mass=0.057 kg) and that I am launching at a 45 degree angle. I assume I would follow the same approach, but alter the h for the gravitational potential energy to be the y component of the launch tube length? $\endgroup$
    – Gara
    Sep 12, 2020 at 11:48
  • $\begingroup$ Depending on whether 1 foot is vertical displacement or the diagonal displacement (i.e. the spring deformation) of the you will need to modify either a) the gravitational energy, or b) the spring energy. If the displacement of the spring is $\Delta x=1 [ft]$, then change $h =\frac{1ft}{\sqrt{2}}$. If $h=1[ft]$ then spring dispacement is $\Delta x=1[ft]\cdot \sqrt{2}$. I'd be bappy to update the answer if you clarified that part. $\endgroup$
    – NMech
    Sep 12, 2020 at 12:41
  • $\begingroup$ Apologies. The 1 ft is the spring displacement. $\endgroup$
    – Gara
    Sep 13, 2020 at 12:24

to enlarge slightly upon NMech's answer: There is no such thing as a true constant-force spring that uses elasticity of its spring material to generate its reaction force.

Springs that behave almost like constant-force springs are designed by taking a spring that is extremely "soft" and preloading it out to some extension which becomes the starting point for further extensions, and then limiting its maximum extension beyond that. The result is a spring that pulls back with almost constant force throughout its extension range- a good example of which is the retraction spring inside a 20' tape measure, where you only pull the tape out a couple of feet.

If you really require constant force to launch a projectile, you can get it by affixing the throwing arm to a wheel, wrapping the wheel with a piece of rope attached to the wheel at one end, and hanging a weight on the other end. Releasing the weight will pull the rope which rotates the wheel, and the torque applied to the throwing arm will be constant throughout the movement of the throwing arm.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.