# BLDC: Required voltage to get set rpm at given amps

1. Mar 9, 2017

### yavorh

Hey guys!

So I'm trying to do this pretty wide diameter in-runner engine, where the rotor is hollow (essentially trying to make a short tube within a short tube :D). After checking out the details between the different types of electric motors (and realizing that I can barely find any instructions on constructing your own AC motor, unlike the instructions to a BLDC motor) I've decided to settle for a BLDC motor.

The set up is as follows - 1.7 meter radius of the stator, probably about 1.65m radius for the rotor. Rotor will be about a 5cm or so thick tube. Thus inner radius of the rotor is about 1.60m, if it's important for you guys.

In any case, while I was calculating the needed voltage and amps I'll need for the motor (thus to calculate the needed battery storage and weight) I was following these two articles for the most part:

http://www.instructables.com/id/Make-Your-Own-Miniature-Electric-Hub-Motor/ For the basic calcs

learningrc.com/motor-kv/ for the Kv, Kt and Ke "ratings" which I would need in order to calculate the needed voltage

Following along these articles, this is the progress that I had made and where my issues arose.

So, taking the formula that T = 4 * m * N * L * R * i , where the values are respectively Torque, constant for BLDC, m- added teeth per phase, loops of wire per tooth, length of the magnets/motor, radius, amps. I had gotten this 101 Newton Meters (what I calced that I need to get my propeller of 3.5kg up to 3200 RPM) = 4 * 33 teeth per phase * 1 turn (6 AWG wire) * 0.02 meters "length" of magnets * 0.825 meters radius of rotor * 50 amps (borrowing from Tesla, calculated them to be running 50 apms and decided to use it as a good baseline...)

At the time I was calculating the needed amps and turns of wire, at the end I settled for 50, as I mentioned above. Here comes where things get poopy. Now that I have my example amps, I need to get the voltage needed in order to start working on my battery set up.

Sooo... the formula given in the second article:
V = R*I + Kt * w (gamma, as in rads per second, can't write it with this keyboard :D)
And the Ke = Kt = T/I0 leads me to the following calculations:
Kt = 101/50 = 2.02 ; 3200 RPM (max rpm, for given torque of my prop) = 335.1 rads/s
V = 0.05184 ohm (20 meters of 6 Awg *2) * 50 + 2.5*335.1 = 840V .. and that's for only ONE of the two propeller-moving motors!

That... seems a bit excessive. Hell, when I was going with lower gauge wire and higher amounts of turns and could get the amps down to 2 it would skyrocket to something about 16KV+

Now, I understand I am moving a wide load, but one would imagine that the amount of teeth, the ginormous radius, small load, etc, would equal to a slightly.... lower sum. Heck, to achieve one hour of flight with this I can't even start to imagine how many what sized batteries I'd need.

I have spoken to some friends who have some long-forgotten knowledge in this area and they told me that they think the formula is wrong, so if anyone can correct me on this or could give me some suggestions, I would be EXTREMELY grateful!

2. Mar 9, 2017

### yavorh

Bump? Wrong section of the forum?

3. Mar 10, 2017

### Baluncore

It is the right forum, but it does not add up.

That is going to be a significant flywheel with gyroscopic inertia.

How will you lubricate the bearings with that surface speed and inertia.
You should attach a sketch of the device showing dimensions.

What RPM do you expect and what diameter do you really need.
Is there a gearbox involved?

4. Mar 10, 2017

### yavorh

Hey Baluncore, thanks for the reply.
The object in question is an airplane propeller. The 3200 RPM is its max required RPM, optimal RPM being 2600. The weight of the propeller(3 blade) is 3.5kg. The propeller is designed to be pushed by a max 60HP engine which is how I calculated how much torque I'd need [60 = (T*3200)/5252)] = ~100 N.m. Approximately 98.5, on my initial calculation I had pulled 101.

I can't mention too much as this is a project that we are developing for patent purposes, but the idea was that instead of having and engine in the middle of the propeller, to instead have a giant in-runner AROUND the prop.

And this is where I ran into the above problems.

5. Mar 10, 2017

### jim hardy

Hmmm ducted fan driven from the periphery... sounds logical, if you can make one that's practical.
Seems to me you'll have to have heavy magnetic material out there at the ring anyhow.
Why stick with a conventional prop? Seems to me a multiblade surrounded by a linear motor would make better use of the magnetic material , with more blade passings you'd get multiple 'pushes' per turn
Then with small magnetic blade tips it becomes a conventional motor. You could probably adapt a washing machine motor controller . Search "Field oriented control"

Navy has some efficient propellers for their submarines but details are sparse.
http://americanhistory.si.edu/subs/anglesdangles/taming2.html

6. Mar 10, 2017

### yavorh

Hey Jim,

The idea is that the propeller, even though it has 3 blades, will be attached to a thrust bearing - which is essentially a full ring which will be absolutely covered in magnets.. Pretty much for those 99 teeth (33 per phase) around the entire diameter of the stator, the rotor (that thrust bearing) will be covered in 200 magnets, each neodynium N42 5kg adhesive magnets.

Also, not sure what you mean by a multiblade surrounded by a linear motor. Could you elaborate?

7. Mar 10, 2017

### jim hardy

click the ∑ symbol in bar at top of posting area and it offers you a list of symbols. Radians per second is usually written as omega, ω, about ¾ along first line of them.

I do not know why your voltage comes out so high.
Will try to look at your references tonite.

8. Mar 10, 2017

### Baluncore

Radius = 1.7 metre. Diam = 3.4 metre. Circumference = 10.7 metre.
3200 RPM = 53.33 rev/sec.
Motor and blade tip velocity = 10.7 * 53.33 = 570.6 metre/sec.
But the speed of sound is only 343. metres/sec.
Mach 1.66 suggests something is very wrong with your parameters.

The blade tips must be thinner as they approach the speed of sound. It will be difficult to support the propeller and transmit 60 HP through such thin tips.
How will you control the armature to stator clearance? I expect you will need to integrate a non-contact levitating magnetic bearing into the propeller tip - motor. That will have significant implications to motor design.

9. Mar 10, 2017

### yavorh

Oh, pfft, sorry... 1.7 meter DIAMETER.
Through the thrust bearing:

The propeller will be attached to one side, the three tips touching one ring, and the other end will be attached to the frame of the airplane. The stator, positioned around the bearing, will be also attached to the frame.

10. Mar 10, 2017

### jim hardy

Wrap a linear motor in a circle and it could pull on one magnet at each blade tip.

That sounds like a good approach. Then you could insert any number of blades.

Those articles both look to be good "How To" practical articles.

Just a sanity check here , always cross check a derived calc against your basics for congruence
50 horsepower at 50 amps requires:
50hp X 746w/hp = 37,300 watts or 37,300 VA , which divided by 50A = 746 volts
i'd say your 840 is reasonable

11. Mar 10, 2017

### yavorh

Hmm... Fair point. What threw me off was looking at Tesla's Model S for reference. As they can achieve up to 500 HP on the rear drive (let's even call it 250 per wheel), with a torque of 400N.m (again, let's even call it maybe 200 per wheel), yet they are rated at 350V at 80kwh.

So, I guess it's a matter of finding the best balance between Volts and Amps so that I can have the least amount of batteries hooked up in parallel and series to power the system.

12. Mar 10, 2017

### Baluncore

Do you really expect to use a ball thrust race at 3200 RPM, with a diameter of 1.7 metres ?
What lubricant ?

13. Mar 10, 2017

### yavorh

Good point. Added to research list for this weekend.

Thanks for the input thus far, guys, it really is awesome and helpful. :)

14. Mar 10, 2017

### jim hardy

hmm once again go back to basics
500hp X 746 watts/hp = 373000 watts = that many N⋅m/sec
932.5 rad/sec / 2π = 148 revolutions/sec = 8904 RPM ,

a two foot diameter tire at 8904 rpm would be going 635mph, too fast for passenger car tires so i dont know quite what to make of their claim. Torque and power aren't measured at same speed?

15. Mar 10, 2017

### yavorh

Exactly!
This is what made me want to double check it with other people, because when I applied the above formulas to both the Tesla example and other given propeller examples on the topic of Kv, Kt and Ke, it came out with wild numbers.

16. Mar 10, 2017

### Baluncore

An off the shelf 500mm diam thrust ball race, with oil has a limiting speed of 560 RPM.
A 1700mm diam race, (if you could get it), would be limited to 165 RPM.
A thrust ball race that would run at 3200 RPM with light oil could only be about 50 mm diameter.

17. Mar 10, 2017

### yavorh

Hmm, yep, just doublechecked the values, the best I can get would be even less - 143 RPM... Well that idea is out the window, back to the drawing board!

18. Mar 10, 2017

### jim hardy

That's the problem i have with formulas
if i don't know exactly what physical thing every single term represents and why its units were selected to whatever they are
i always miss a conversion factor and get crazy answers.

Your project is really interesting, moving the airplane motor to inside a NACA cowl and making it a ducted fan.

applying the torque way out there reduces necessary electromagnetic force
but moves significant mass out to periphery , which will give that prop a big gyroscopic effect.
That's interesting because a single phase coil produces two counter-rotating magnetic fields which might lend itself to counter-rotating props

I'd be in way over my head . I'd have to start with force i need to apply to the rotor, from that figure magnetic field strength needed, from that the cross section of iron and number of amp-turns to push that much magnetic flux through the iron and the air gap,
then look at whether i could make a structure rigid enough to support the mass of my rotating magnetic ring using blades as spokes instead of a thrust bearing.

Might work out that a small high speed motor with planetary reduction gears driving from center would have the weight advantage.

Lessee here.......

50 hp on 1.7 meter diameter at 3200 RPM.....

hp = 2π X torqueft-lbs X rpm / 33000
torqueft-lbs = hp / (2π X rpm / 33000 ) = 50 X33,000 / (2π X 3200) = 82.06 ft-lbs = 111.3 N⋅m

111.3 N⋅m on a 1.7 m diameter is tangential force of 96.55 Newtons?

Found by searching on terms 'tangential force on motor rotor' an online book on the subject
here's a snip from it

A is current density which is one way to express the strength of permanent magnets.

That's a basic principles approach against which you should be able to cross check those formulas from your RC motor books.
That book answers an old question in another thread about effect of slots on armature conductors that's had me stumped, even my old college prof didn't have a good answer. Thanks for causing that serendipitous discovery!

Good Luck, Guys !

old jim

Last edited: Mar 10, 2017
19. Mar 10, 2017

### yavorh

Daaang, you certainly can put down the hammer when it comes to information gathering.

It is a shame that this idea will need to be put on the wayside, at least for this application of such a rotor, but I will definitely have this in mind in case we manage to find another application of it.. Low rpm, high torque applications, perhaps? One can only wonder...

By the way, any idea what the B in this formula may refer to?

20. Mar 10, 2017

### jim hardy

Surely it's flux density
i didn't read far enough to see if it's SI Teslas or cgs Gauss .