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Old Nov 23, 2003, 03:23 AM
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Actuator calculation spreadsheet.

Hi

I have made the actutator calculation spreadsheet from Gerhard Ringel available for download. Here is a direct link:

http://www.tekmek.dk/copper.xls

Here is Gerhards instructions for use:

"Here is the spreadsheet. to use, just put in the inside diameter of the coil into H2, the outside diameter into I2, the length (width) of the coil into J2, it will show you the weight of the (copper in the) coil in K2, and the average length of each turn in L2.
You can then determine the Ohms and number of turns form your wire size. I have listed the common American wire gauges and a few wires by mm thickness."

"I have added a formula as you suggested on page 2. You can enter the wanted Ohms in M2, the inside diameter in N2, and the width of the coil in O2. It will calculate the total length of the wire, the Outside diameter of the coil, the number of turns required, and the weight of the copper in the coil.

Note that all this is based on just the copper. Actual wire will include the lacquer insulation, so the actual OD and the actual weight will be a little more.

To get the values for other wire sizes, you can input the diameter of the wire in mm in any of C29 or C30 or C31 or C32 and it will do all the calculations for that wire size.

I am not familiar with the standard wire sizes for metric.

Of course I would be happy if you could post this information on your web site. I want this to be available to as many as want to see it.

I have also added a column for the maximum current for a give wire size. I came up with a formula based on various information I could find on the web and some experiments. This is approximate and some folks may want to be a little more aggressive, some a little more conservative."

Thanks a lot to Gerhard for sharing this spreadsheet If anyone has suggestions for improvement or would like to have additional features, just post it here.

Graham or Gordon, please add this thread to the Useful Threads.

Michael
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Old Nov 23, 2003, 06:24 AM
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Thanks for posting this Michael, I was actually working on something similar because almost every wire chart that I found was a little different.

I was having trouble finding suitable magnet wire, and was going to go ahead and order some 42.5 AWG (smallest they have) from http://www.brigarelectronics.com/ until I found out a 200 Ohm coil with this size magnet wire would weigh almost a gram!

The best one I found so far is based on NEMA standards for magnet wire.

www.elektrisola.com/nema.xls

It has a lot of other info like MIN, NOM and MAX for bare copper wire, single build, heavy build and triple build magnet wire as well as resistances and winding tensions.

It has sizes from 24 AWG to 44.5 AWG. It has higher sizes but says "45-58 AWG Theoratical bare wire diameter by resistance".

Thanks!
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Old Nov 23, 2003, 03:13 PM
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Dindea
You can get some wire from Wally World. The cheapest single setting timer has enough for 15 - 200 ohm coils. I can send you some for a few actuators.

Michael
Thanks for the table.

Billy
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Old Nov 23, 2003, 03:34 PM
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Maybe we could add a torque estimate somehow if the same weight magnet as wire is used. Hasnt someone stated a formula for that? Maybe it depends on magnet strength though.

Billy
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Old Nov 24, 2003, 06:44 PM
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Great info

Thanks all for great information.

John
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Old Nov 24, 2003, 06:46 PM
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Quote:
Originally posted by billystiltner

Michael
Thanks for the table.

Billy
All credit goes to Gerhard. I just put the spreadsheet on my webspace.

It is useful though, isn't it?

Michael
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Old Nov 24, 2003, 08:03 PM
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I'll rephrase. Thanks for sharing Gehard's table with us Michael.
It is very useful. I did know that though. Looks like I can wind a 200 ohm coil with .03mm wire at around .05g if then I add a magnet that weighs around .05g, I will have a close to .1g actuator if my bobbin or straw weighs next to nothing. This is good news. I bet I could even make 2 close to that weight with thinner wire and smaller magnets. Let me check.

Billy
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Old Nov 25, 2003, 09:24 AM
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Michael,

Thank you for posting the spreadsheet! ( I would not have known how)

Billy,
If you make an actuator that small, you may not get any useful torque out of it. In the spreadsheet, on page one, there is a hidden column, column "L". You can see it if you select cloumns K and Q, right click the mouse and then select 'unhide'. This column will give you an indication of relative torque.
By measurement, I found that with a NdFeB magnet of about 200mg, a value of 40,000 in column L gives a torque of about 1.4 gcm.
Using the values you gave above, with a 50mg magnet, and assuming you can make the coil inside diameter 4mm, your expected torque would be less than 0.1gcm.
Is that enough?

Maybe I'll add a column with the expected torque based on the weight of the magnet.

regards,
Gerhard

P.S. you must have very small fingers!
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Old Nov 25, 2003, 11:05 AM
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Gerhard
Probably so I have no grasp on the torque figures yet.
It will be used on planes 6" - 8" span weighing from 5 to 9g.
Anywhere from 12 " ^2 area or less on up.

I have heard of actuators this small being able to control small planes so I am only guessing it will be enough.

I tried unhiding the L column but it did not work.
I'll download again.

Thanks

Billy
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Old Nov 25, 2003, 11:12 AM
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I got it to work
for some reason you have to click shift to get it to unhide.

Billy
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Old Nov 25, 2003, 08:11 PM
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This spreadsheet is great.
I used to wind a 100 or 50 turns or so then
measure resistance and then calculate how many turns
I would have to do to get the desired resistance.
This is a lot better the calculation is done in the spreadsheet.

Thanks again.

Billy
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Old Nov 25, 2003, 08:24 PM
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Thanks Michael and Gerhard for making this available

I may be wrong but the equation for number of turns seems to be incorrect. In does not include the length of the coil as a variable, how is this possible?

I also see you have calculated the OD by using the volume of the copper required for the wanted resistance and then effectively assumed the wire has a square cross-section to work out the OD. Any idea how much of an error this introduces?

I think this is a really handy idea. I tried to do similar last year and did manage to make an errorlessish equation for the OD and number of turns but I seem to remember there had to be full layers of the coil for it to mean anything. Ah, perhaps I was calculating R from the number of turns. Anyway it involved thinking of the coil as a really long cone. The additional diameter you get per layer was just spread out as a steady gradient. The point is it was a little useless and I decided a proper computer program would be needed to get the results correctly, at least with my mathematics skills.

Thanks

Graham
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Old Nov 25, 2003, 09:18 PM
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Hi Graham,

I call it the width of the coil, you call it the length, it has to be entered in J2. It is definitely used in the formulas.

You are right about my assumption that the entire volume of the coil would be occupied by copper. It does not introduce much of an error for the weight, only for the actual outside diameter of the coil. That is typically not critical.
Actually, think of the wire as being hexagonal. Except for the edges, it will pack in completely. a hexagonal wire with the same small diameter (measured from face to face) as the round wire, packs in the same as the round wire, but fills up all the volume in the coil. the difference in cross section beween them is as 2*sqrt(3) to PI, or about 1:1.10, 10 percent. So the actual volume of the coil will consist of 91% copper, and 9% air, so the outside diameter will be about 4.5% larger than expected from a completely packed coil.
Another source of error is not accounting for the weight of the laquer or enamel insulation. I do not think it weighs more than 5% of the total, maybe much less, but I am not sure.

to get the total length of the coil, I just assume that all the turns are the same as the average based on the average diameter of the coil (ID+OD/2). This is actually accurate. Trust me, I have a Master's degree in Math.

Even if there is a small error in the outside diameter, the error is small, probably less than the manufacturing tolerance of the wire.

regards,
Gerhard
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Old Nov 25, 2003, 09:22 PM
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BTW,

The most accurate methode of getting the wire size I found is to take a known length of wire, putting a known current thru it, and measuring the voltage drop. you can then get the Ohms per meter, and look it up in the table.

But you already knew that, huh.

regards,
gerhard
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Old Nov 25, 2003, 09:27 PM
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Wouldn't it be (inner circumference + outer circumference) /2?
I may be wrong here but for a 6mm ID x 4mm wide coil wound with 652 turns of .03mm wire should be about 200 ohms. I will have to measure the AWG or m to make sure but this is what I wound and I got 200 ohms I think it is from memory so I will go look it up and measure.

Billy
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