Calculating spring/motor requirement

[blixel]

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:

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?

 

[Dr.D]

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.

 

[blixel]

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.

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.

 

[Dr.D]

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.

 

[blixel]

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.

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.

 

[JBA]

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.

 

[blixel]

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.

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.

 

[blixel]

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

 

[Dr.D]

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.

 

[blixel]

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.

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.

 

[Dr.D]

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.

 

[blixel]

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.

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.

 

[Dr.D]

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.

 

[blixel]

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.

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.

 

[JBA]

,What are the units of the spring constant values in your table?

 

[JBA]

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.

 

[blixel]

,What are the units of the spring constant values in your table?

joules/m^2

 

[JBA]

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.

 

[Dr.D]

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.

 

[blixel]

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.

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?

 

[Dr.D]

If you work in USC with length in feet, the energy will be in ft-lb and the spring constant will be in lb/ft.
If you work in USC with length in inches, the energy will be in in-lb and the spring constant will be in lb/in.

 

[blixel]

If you work in USC with length in feet, the energy will be in ft-lb and the spring constant will be in lb/ft.
If you work in USC with length in inches, the energy will be in in-lb and the spring constant will be in lb/in.

Hmm... I can't follow how the units cancel to get to ft-lb for energy.

 

[JBA]

For 45°:
If W = .0585kg *2.205 = .129 lbs
US mass = W/g = .129/386.16 = 3.34E-04
KE = .5 m v^2 =.5 * 3.34E-04 * 177.165 (in/sec) ^2 = 5.241
KE = PE sprg = 1/2 k x^2
k = 2 * KE / x^2
For x =4 in:
k = 2 * 5.241 / 16 = 0.655 lb/in
F = k * x = .655 * 4 = 2.62 lb

PS in US units my tennis ball weights 2 oz = .125 lb

 

[Dr.D]

Hmm... I can't follow how the units cancel to get to ft-lb for energy.

If you work in USC with length in feet, then mass is in slugs = lb-s^2/ft. You did your calculation with weight rather than mass. This is a common mistake, but it will consistently get wrong results.

This is equivalent to doing the calculation in SI with mass expressed in Newtons. Most folks know that is incorrect, but for some strange reason, they think that in USC, mass is in pounds. Why is that? Do we not understand weight in USC?

 

[blixel]

For 45°:
If W = .0585kg *2.205 = .129 lbs
US mass = W/g = .129/386.16 = 3.34E-04
KE = .5 m v^2 =.5 * 3.34E-04 * 177.165 (in/sec) ^2 = 5.241
KE = PE sprg = 1/2 k x^2
k = 2 * KE / x^2
For x =4 in:
k = 2 * 5.241 / 16 = 0.655 lb/in
F = k * x = .655 * 4 = 2.62 lb

PS in US units my tennis ball weights 2 oz = .125 lb

This part "US mass = W/g = .129/386.16 = 3.34E-04" is completely out of left field for me. I get where W=0.129 lbs, but what is "g" in this case? It doesn't appear to be gravity (32 ft/s^2).

Regarding the tennis ball mass and diameter, I was just going by:

 

[JBA]

g is 32.18 ft/sec * 12 in/ft = 386.16 in/sec

 

[blixel]

If you work in USC with length in feet, then mass is in slugs = lb-s^2/ft. You did your calculation with weight rather than mass. This is a common mistake, but it will consistently get wrong results.

This is equivalent to doing the calculation in SI with mass expressed in Newtons. Most folks know that is incorrect, but for some strange reason, they think that in USC, mass is in pounds. Why is that? Do we not understand weight in USC?

(No wonder we use SI.) At any rate, is there a preferred system here that I should be using for the sake of this problem? Recall that I'm trying to create calculations that I can use to eventually buy a spring and a motor in the USA.

 

[blixel]

g is 32.18 ft/sec * 12 in/ft = 386.16 in/sec

Got it. But why are we suddenly changing from ft/sec to in/sec here when all the original work to calculate velocity was in ft/sec?

 

[JBA]

Apparently now US spring manufacturers now allow both US and SI units to be used, below is an online catalog for Lee Springs, who I have used for years for off-the-shelf spring purchases. Just click on the style of spring you want and you will see how to do a selection using either US or SI units.

http://www.leespring.com/

As a retired american engineer who is required to use SI for many problems on this forum, SI is just as frustrating for me as US is for you.

 

[JBA]

To maintain units consistency with the in/sec for the ball velocity and in. for the spring displacement and so I could get the spring constant value in standard US lb/in rather than a nonstandard lb/ft.

 

[Dr.D]

I usually use the acceleration of gravity in USC as
g = 32.174 ft/s^2 = 386.088 in/s^2
I would never consider 32 ft/s^2 as a reasonable approximation to g, particularly when you are going to do the calculations to 8+ digits in a digital machine.

The USC system allows two options. You can use length in feet, or you can use length in inches, but you make a choice and then stick with it.

Whether you choose feet or inches depends upon the scale of your problem. Are the dimensions small? If so, inches make a good choice. If the dimensions are large, the feet make a better choice. Machinery problems are almost always done in inches, even when hundreds to a few thousand inches are involved (I have worked on large diesel gensets where all the dimensions were in inches.).

Most off the shelf springs will use inches for the displacement dimension if using USC. If you want to work in feet, then plan to calculate K in lb/ft and later convert this to lb/in.

It really is not hard if you have a clear understanding of the distinction between weight and mass and know how to work with units. Neither system is perfect, and really, neither system is better than the other. (Strict SI requires all lengths in meters, not mm or cm. Try expressing the thickness of sheet metal in meters.) There are some awkward sizes in each system; that's just life!

I do not find SI at all hard to work with, although I have much less feel for the numbers in SI. When a stress is 153 MPa, is that a lot or a little? If it was in PSI, I would have a lot more feel for it, but that is simply the result of many years working as an American engineer.

 

[blixel]

At 45 degrees, I seem to be getting a different velocity than you. I'm not sure why that is, but assuming the calculations are correct otherwise, this image is where I'm at now on the process. Though I am definitely losing track of units at this point.

• Initial Velocity is in ft/s.
• Then Kinetic Energy is ... ft-lb (I think), or am I now in slugs, or ft-in, or ...?
• And the Spring Constant is now lbs/in I believe.
• And finally the Force is just pounds?

 

[JBA]

I used 4.5 m/s * 39.37 in/m = 177.165 in/sec

 

[JBA]

Our velocity values don't differ by much, my 177.165 in/sec converts to 14.76 ft/sec as opposed to your 14.72 ft/sec.

 

[Dr.D]

I see that in your spread sheet, you show the mass in pounds. This simply will not work right.

g is the acceleration of gravity (this is almost universal usage).

If you use feet for length,
then
velocity is in ft/s
acceleration is in ft/s^2
weight is in lb
mass is in slugs = lb-s^2/ft
energy/work is in ft-lb
g=32.174 ft/s^2

If you use inches for length,
then
velocity is in inches/s
acceleration is in in/s^2
weight is in lb
mass is in lb-s^2/in
energy/work is in inch-lb
g = 386.088 in/s^2

The key is be consistent.

Let's talk philosophy for a moment to see if we can clarify this whole matter. Newton's Second Law reads
F = M*a
On the left, we have a force, with units that we will denotes as F.
On the right, we have a product of mass and acceleration, with units M*L/T^2
where
F - force unit (includes weight which is a force)
M - mass unit
L - length unit
T - time unit
This indicates that there are four different types of units that must be related to each other by one constraint (Newton's 2nd Law).
Three of the four can be chosen arbitrarily, but the fourth is determined by the constraint.

In the SI System, the three fundamental units are
mass (kg), length (m), and time (s)

In the USC System, the three fundamental units are
force (lb), length (ft or in), and time (s)

Notice the fundamental difference between the two systems. SI is "mass based," while USC is "force based." If mass is fundamental, the force becomes a derived unit. If force is fundamental, then mass becomes a derived unit. Each system is consistent within itself, but they are not set up quite the same at the fundamental level.