Calculating spring/motor requirement

[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.

 

[blixel]

there is a Lowe's by where I live so I came by to look at their springs so I would know what values they are using. Their springs are listed just by their safe working load in pounds. Is that the same as lbs/in?

 

[Dr.D]

No, the safe working load is an indication of the maximum load that the manufacturer thinks the spring will support without failing. It tells you exactly nothing about the spring rate.

My guess is that you will have to buy some springs and determine spring rates experimentally. It is not hard, but just takes a little bit of lab work.

 

[JBA]

To get a general sense of the springs that be in your general requirements range use the LEE springs site click on the type of spring you want and use the selector by entering your desired rate, and load and P2 position values. There definite ranges of spring diameters and wire sizes that determine a spring that can meet a particular set of load and deflection combinations and these are controlled by the stress limits of the spring materials available.

If your are looking for extension type springs, given your low rate and long extension requirements, then try looking at hardware stores for the old style screen door springs (if they still exist) these springs are very long and of a relatively low spring constant. Even if one doesn't exactly meet your requirements you may be able to get near what you want by cutting the spring to a shorter length because this will reduce the number of coils in the spring and raise its spring constant (also known as spring rate) and load at a given extension distance. A quick look on Google shows that Home Depot and Lowes both have some springs of this type but those springs, or some of them (like many things theses days) may only be available from the stores websites but not in the stores (you will just have to do some investigating) and good old hands on experimenting. Ultimately, there is number of spring rate and pull back distance combinations that can give you something you can use.

At some point you may find you need to order a spring from a company like Lee Spring to fit your needs. Using the rate and 4 in extension in my calculation I found one spring at Lee Spring that met those specs and cost about $8 for a quantity of one plus shipping of course.

In every case there are tolerance variations on all commercial springs so you will need to include a provision for load adjustment in your machine design.

 

[blixel]

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

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

The key is be consistent.

Trying again...

Does this much seem valid?

 

[Dr.D]

A block with a weight of 1 lb has a mass of M = W/g = 1/32.174 = 0.031081 slugs.
If the item weight is W = 0.128906 lb, the mass is M = W/g = 0.128906/32.174 = 4.006527*10^(-3) slug
Now, look at your units on the KE:
KE [=] M*(L/T)^2 [=] (M*L/T^2)*(L) [=] lb-ft
where
[=] means "is dimensionally equal to"
Without checking your final arithmetic, yes, the last result above, KE=0.3084 ft-lb looks correct. (note that the order of the units makes no difference at all, so that ft-lb is equivalent to lb-ft)

 

[blixel]

A block with a weight of 1 lb has a mass of M = W/g = 1/32.174 = 0.031081 slugs.
If the item weight is W = 0.128906 lb, the mass is M = W/g = 0.128906/32.174 = 4.006527*10^(-3) slug
Now, look at your units on the KE:
KE [=] M*(L/T)^2 [=] (M*L/T^2)*(L) [=] lb-ft
where
[=] means "is dimensionally equal to"
Without checking your final arithmetic, yes, the last result above, KE=0.3084 ft-lb looks correct. (note that the order of the units makes no difference at all, so that ft-lb is equivalent to lb-ft)

If the steps are valid, then I trust the spreadsheet to handle the arithmetic. And I walked through the steps to see how the units cancel and become ft-lb (or lb-ft). So that is also clear to me now. Thanks for that.

As far as going to the next step of calculating the Spring Constant, that should be unchanged... so the last 2 steps for getting k and getting F are...

I feel like this is definitely correct... but my head is spinning so I want to check.

 

[Dr.D]

I have not checked your arithmetic, but it looks OK.

 

[Dr.D]

I was thinking earlier, however, that you may have used the wrong mass value. You based your mass on the weight of the projectile only, but it should include the "chair" and (1/3) the total weight of the spring. This will modify your calcs above by a little bit, but not too much. The spring weight is, of course, initially unknown. You can start with an estimate, and iterate the calculation with closer and closer values as you learn more.

 

[blixel]

I was thinking earlier, however, that you may have used the wrong mass value. You based your mass on the weight of the projectile only, but it should include the "chair" and (1/3) the total weight of the spring. This will modify your calcs above by a little bit, but not too much. The spring weight is, of course, initially unknown. You can start with an estimate, and iterate the calculation with closer and closer values as you learn more.

Ah ... so the mass should be (tennis ball + chair + 1/3 of the spring). I'll definitely keep that in mind once we get done with the paper and pencil portion and start the actual build. That will be an easy adjustment for me to make on my end. Since I'm using the spreadsheet, and since the calculations won't change or have to be updated ... I'll just add a couple of cells for the chair and 1/3 of the spring, and then have the mass calculation be a sum of the 3 as opposed to just the tennis ball. (And remembering that when I input the "masses" for the chair and 1/3 of the spring, I'll actually be putting them in as Weight, which will then be converted to slugs.)

 

[blixel]

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.

ha, yeah - I understand. I think in my case it's a matter of wanting to solve, what seems to be, a reasonably simple problem. But being inundated with so much extra information that it makes me want to give up on the math, and just figure it out through experiment.

At any rate, the main reason I'm replying is to ask about that website you linked to. I go there, and I believe the type of spring I'll want is an Extension Spring. One end will be nailed/screwed/secured to the launch point, and the other will be secured to our chair. BTW, this video (below) showing what this guy built is what inspired our design. We are trying to build the same thing (more or less).

But getting back to leespring, do you happen to have a direct link to a spring meeting the general requirements? The site isn't terribly intuitive to someone with no prior experience in buying springs.

 

[JBA]

Honestly, it is not very intuitive for me as well. In the past I have been able to simply leaf through a copy of the catalog to search for a close match to my requirements but this online catlog makes that approach totally impractical; so, below are is the input I used and the result based upon my #23 post solution for the 45° case. You bring up the extension search app by clicking on the picture of the extension spring in the list of spring types on the left side of the page.

 

[blixel]

Honestly, it is not very intuitive for me as well. In the past I have been able to simply leaf through a copy of the catalog to search for a close match to my requirements but this online catlog makes that approach totally impractical; so, below are is the input I used and the result based upon my #23 post solution for the 45° case. You bring up the extension search app by clicking on the picture of the extension spring in the list of spring types on the left side of the page.

View attachment 114667 View attachment 114668

Ok, I'm able to repeat your values and get a list, but if I change just about anything, there are no results. After holding some springs in my hand at the store, and stretching them a bit, my feeling was that 4 inches isn't enough. So I plugged 6 inches into the spreadsheet to generate a new set of values, and am getting no results from the website. I tried 8 and 9 inches also, but no search results. Odd.

 

[JBA]

When I run it I get a "no items found" result regardless of load or extension so that I suspect that the required wire size for that very low constant is too small. I did a quick catalog search on their smaller springs and the lowest rate is .16 on even very small springs with a .007 inch diameter wire size as you can see in the lower section of the below screen print of that section of their catalog.

 

[blixel]

Continuing on with this ball launcher project... nearing the completion date, so one way or the other, it will be over in about 10 more days.

The motor I got was https://www.amazon.com/gp/product/B009YES7EO/?tag=pfamazon01-20 motor from Amazon. According to Page 8 of the manual, it has a maximum Torque of 420 in-lb. I really don't know how to translate my earlier spring constant and force calculations into Torque, but I was hoping that this motor was substantially stronger than was actually needed. At any rate, the big issue is that it spins extremely fast. So my question for the forum now is ... does anyone know where I can get a large diameter gear that I can drive with this motor? The idea being that the smaller motor gear will spin the larger gear and lower the RPM while also creating more torque.

Alternatively, if I can find a different motor that is designed for low RPM and high torque, I can just replace this one. Bearing in mind that the project has a tight budget, so any motor I get would have to be in the same low price range, with shipping cost included. (So Amazon prime works best for me because shipping costs come out of the project budget.)

I was reading someone else's post about stepper motors on an Arduino forum. That sounded promising but they were talking about additional controller boards that were needed to drive the stepper motor. I'm worried that that would get too expensive, and might require programming and software ... which I don't feel like I have time for at this point.

Ideally I could find a motor that could just be turned on and off by interrupting the power (like the one I have), but is strong and doesn't have more than about 80-90 RPM.

Any input is appreciated.

 

[JBA]

The minimum torque required will be your design spring load at it initial installed deflection x the radius of the motor shaft gear and the maximum torque required is the spring load at full design deflection times the motor gear radius.

ALERT: WHAT YOU HAVE HAVE SELECTED IS A CRAFTSMAN 11543 HAMMER DRILL. ARE YOU SURE YOU WANT HAT TYPE OF DRIVE? UNLESS YOU ARE VERY FAMILIAR WITH AND HAVE USED THIS TYPE OF DRILL, I WOULD NOT THINK SO; NOR, WOULD I RECOMMEND THAT TYPE OF DRILL FOR YOUR APPLICATION.

Apart from the function issue of the hammer action of this type of drill, I would expect that its torque far exceeds your requirements i.e. at its rated torque of 420 in. lb., if you put a 1 inch diameter gear or pulley on its shaft, it will be able to pull a maximum of 420 x .5 = 210 lbs of force which I am sure far exceeds the maximum spring load of your device.

At the same time, it does have an adjustable speed and you can get a standard variable speed hand drill with that function; but, I would think that any of those will also far exceed your max load requirement.

 

[blixel]

The minimum torque required will be your design spring load at it initial installed deflection x the radius of the motor shaft gear and the maximum torque required is the spring load at full design deflection times the motor gear radius.

ALERT: WHAT YOU HAVE HAVE SELECTED IS A CRAFTSMAN 11543 HAMMER DRILL. ARE YOU SURE YOU WANT HAT TYPE OF DRIVE? UNLESS YOU ARE VERY FAMILIAR WITH AND HAVE USED THIS TYPE OF DRILL, I WOULD NOT THINK SO; NOR, WOULD I RECOMMEND THAT TYPE OF DRILL FOR YOUR APPLICATION.

Well, when I connect the motor to a DC power supply (like a 6V lead acid battery or DC variable power supply), the motor just spins as I would expect any motor to. It just spins way too fast for my task.

Apart from the function issue of the hammer action of this type of drill, I would expect that its torque far exceeds your requirements i.e. at its rated torque of 420 in. lb., if you put a 1 inch diameter gear or pulley on its shaft, it will be able to pull a maximum of 420 x .5 = 210 lbs of force which I am sure far exceeds the maximum spring load of your device.

At the same time, it does have an adjustable speed and you can get a standard variable speed hand drill with that function; but, I would think that any of those will also far exceed your max load requirement.

Yes, I do think 210 lbs of force is probably more than we need, but if you refer to the video below, I don't think being over powered is an issue. Once the latch goes past the end of the belt, it just releases the spring. But I do need to be able to slow it down. I don't need more than a few dozen RPM at most.

 

[JBA]

An inexpensive standard adjustable speed battery powered hand drill's speed can be adjusted all the way from 1 rpm to it maximum rated speed; and, if you are using the above type of drive then having a excess of torque available is not an issue.

Look for that type of drill and you will find one much less expensive than the one you specified above.

Below is an example of the type of drill I am describing.

http://www.homedepot.com/p/BLACK-DE...PIPHorizontal1_rr-_-205152814-_-202891148-_-N

 

[JBA]

As an adendum on the torque requirement issue, the best place to begin addressing this is to establish the diameter of the sprocket wheel you want to use for you chain drive and multiply that times your maximum spring load and work backwards from that.

 

[blixel]

An inexpensive standard adjustable speed battery powered hand drill's speed can be adjusted all the way from 1 rpm to it maximum rated speed; and, if you are using the above type of drive then having a excess of torque available is not an issue.

Look for that type of drill and you will find one much less expensive than the one you specified above.

Below is an example of the type of drill I am describing.

http://www.homedepot.com/p/BLACK-DE...PIPHorizontal1_rr-_-205152814-_-202891148-_-N

So if I understand this whole thing correctly, the motor I got, plus a https://www.amazon.com/gp/product/B00UMK4NKG/?tag=pfamazon01-20 would allow me to control the speed of the motor?

Though, it does seem like the link you posted is a better value as you get all the parts in one + battery and charger. But am I correct in thinking that the speed controller would accomplish the same thing?

 

[blixel]

As an adendum on the torque requirement issue, the best place to begin addressing this is to establish the diameter of the sprocket wheel you want to use for you chain drive and multiply that times your maximum spring load and work backwards from that.

Yeah, this is also an area of the build that is proving problematic. Once the motor (and speed) are solved, then there is the issue of figuring out where to get a matching gear and belt.

 

[JBA]

So if I understand this whole thing correctly, the motor I got, plus a https://www.amazon.com/gp/product/B00UMK4NKG/?tag=pfamazon01-20 would allow me to control the speed of the motor?

I wasn't under the impression that you already had a motor, I thought you were considering getting the drill you referenced.

 

[blixel]

I wasn't under the impression that you already had a motor, I thought you were considering getting the drill you referenced.

In message 49, I mentioned that I already got the motor. But later in the same post, I did say that I was open to the idea of finding an alternative. For the sake of the project budget, I'm only including the costs for the items that actually go into the finished project. If I end up with extra stuff, I'll return what I can or just use it for some other purpose.

 

[JBA]

I missed the "I got" part of that post statement.
What you keep describing as a motor, because of your application is in reality for the outside world known generally as a "Hammer Drill" and you should use that name for it in your posts so that other do not confuse it with simple electric motors.

In that respect, the controller you are looking at is not suitable to be used with a battery operated hand drill or hammer drill. That controller is meant to be used with standard DC electric "motors".

At the same time , what you are overlooking in the spec sheet for your purchased hammer drill is that it is a variable speed unit all within itself. If you will look at the speed listing on that spec sheet you will see:

Never the less do you really know how a "Hammer Drill" works? They specifically designed for drilling into hard materials using carbide tipped special drills for that sole purpose like concrete so as soon as they come under torque they start a hammering action along with their rotation, like someone hammering the drill into the hard material with a hammer. They do not deliver the smooth rotation that you need for your mechanism.

 

[blixel]

I missed the "I got" part of that post statement.
What you keep describing as a motor, because of your application is in reality for the outside world known generally as a "Hammer Drill" and you should use that name for it in your posts so that other do not confuse it with simple electric motors.

In that respect, the controller you are looking at is not suitable to be used with a battery operated hand drill or hammer drill. That controller is meant to be used with standard DC electric "motors".

At the same time , what you are overlooking in the spec sheet for your purchased hammer drill is that it is a variable speed unit all within itself. If you will look at the speed listing on that spec sheet you will see:

View attachment 115102

Never the less do you really know how a "Hammer Drill" works? They specifically designed for drilling into hard materials using carbide tipped special drills for that sole purpose like concrete so as soon as they come under torque they start a hammering action along with their rotation, like someone hammering the drill into the hard material with a hammer. They do not deliver the smooth rotation that you need for your mechanism.

Ok I was not aware of that. I will definitely have to find a different motor.

 

[blixel]

Well the project is finally over. Yesterday was demonstration day. The design my group chose proved to be too complicated and we weren't able to get a working prototype in time. However, we still demonstrated what we had completed.

At any rate, here are some short (<1 minute) video clips of the teams who did complete a working prototype. Though, only 1 of them was able to hit the 6 to 10 feet target. (I was only able to take video of the other teams. But our demonstration was completely uneventful since we didn't have a finished prototype.)

Team A hit the target twice (per the requirement)


Team C overshot


Team D undershot


The end.