Tesla's Charging Circuit and it's Application to Pulse Motors

[Farmhand]

MileHigh could you explain what the letters all mean in that formula ?

I = 2MGh/w2 .  2M I see as 2x mass whats the Gh/w2 mean ? I did not lean calculus ever. I'll have to learn as I go. If I have the formula and it expanded explanation of what all the letters mean and the calculating procedure I can work it out.  ie. from memory I think some times the squaring is done first then the multiplying, or does that require brackets.

It's kinda like asking a race car driver to calculate the physical characteristics of his car, that's the engineers job. He would need to sit classes to learn how to do it and that would interfere with his driving practice. Although any good driver knows by intuition what is happening, they just can't explain it in engineer language because they never learned the language. That's why you see drivers mover their hands and bodies in certain ways to explain things of reality to engineers. And if given the tools and time a driver can build a very good race car because he knows what he needs to go fast and what is benfecial to his driving style, if not he's not a good driver. The best professional crews are ones that can communicate properly so the engineers can make happen what the driver wants or needs, while having talent at the same time. One without the other is less than optimal. And people can spread themselves too thin or suffer burn out from taking too much on themselves.

Imagine a motorcycle rider trying to explain the physics behind how he can ride on his back wheel at 160 kph and take corners pass cars ect. He can't  explain it but he can do it. The engineer can explain but can't do it, I know which is more fun and exciting. 

Basically I struggle with the calculations because it's all new I have almost zero experience with calculus. But I can wheel stand powerful motor cycles and drive cars fast and sideways.  I did give it up after having a child and some accidents though.

When doing it it's a kind of melding of man and machine, pure understanding, no thinking required. The thinking usually happens after, sometimes before. I feel the same thing with my electronic devices, I get into the machine kind of thing and "get the feel", use and understanding not on a purely electrical level.

That being said, without engineers the world would be chaos, roofs would cave in regularly, the power would go out all the time ect. I have a lot of respect for educated learning.
It's vital for some to take it on. I did but not in this field. And the education I was provided was only just enough to ply my trade.

Cheers

P.S. I honestly can't even work out if a long run down is better or worse for efficiency, the way i see it a long run down with a large input power from a low speed would be bad.

A short run down from a high speed with a small input would be good.

I can make different results repeatable with the same setup adjusted differently, so something is not right and there is a fudge factor at play. If I was underhanded I would just make it work so it worked out good for me. But I seek the truth not what looks good for me.

I wouldn't discount others fudging results like this so it looks better for them. Too many contributing factors.

I just need to determine how much work the rotor can output as work done, like a dyno setup.

The idea with the small generator as a dyno is good if a direct drive is used and the base load of the generator taken into account, Ohms law could give raw data.

Any dyno with a belt, like a pony brake suffers from friction value changes due to heat in my opinion.

There must be an easier and more accurate way to measure the work output done by the shaft without the parasitic load of a generator, or fudge factors at play.

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[MileHigh]

Farmhand,

Note there is no calculus here, just a straight multiplication and a division, and you do the squaring in the denominator first before you do the division.

You are correct and "M" stands for the mass in kilograms of the weight.

The "G" is probably the wrong case, I think that they use "g."  That's the acceleration due to gravity constant and people normally use 9.8 meters per second-squared.  You can look up the value if you want for more precision.  It's the average acceleration due to gravity on the Earth at sea level.

Note when you multiply "M" and "g" together you get the SI unit for force, the Newton.  The Newton is the real unit for force that's in common use for scientific calculations.

"h" is the height that the weight falls in meters.

"w2" is really "w^2" or "omega-squared."  Omega is the lower-case Greek letter that's used for the angular velocity of the rotor in radians per second.  The Greek letter omega looks a lot like a "w."  A radian is a measure of angle and it's about 57 degrees.   To calculate the angular velocity in radians per second you simply take your frequency in Hertz and multiply by 2 X Pi.   So, omega = (2 x 3.14159 x measured frequency in Hertz.)   In the common short form you say, "omega = 2 x Pi x f"   So one rotation per second is 6.28 radians per second.

The "omega-squared" is calculated first.  There is no squaring of anything in the numerator.

MileHigh

http://www.youtube.com/watch?v=JHhoHmiiXdc
http://www.youtube.com/watch?v=V--pS7Is4P8

[Farmhand]

OK Thanks I'll try to digest that and perform a trial calculation with arbitrary figures.

I think your method will give a valid result if certain procedures to gather the raw data are followed, but I'm not sure what they should be just yet. I just don't want to overlook anything and make a big boo boo.

Much appreciated.

[Farmhand]

When I think about it both methods are the same but just done a bit differently, so a valid result should be assured if it's done correctly and calculated without error. I'm still excited ! 

The way I see it for weighing the rotor parts I could do it like as follows, with reference to the links..

Link to the Engineering Toolbox "Flywheel Kinetic Energy"
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html

Flywheel shapes
http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html

1) the magnets can be calculated in total ( added together) as a hollow cylinder.
2) the squirrel cages can be calculated as solid cylinders
3) the plastic rotor plate can be calculated as a solid cylinder or maybe a hollow cylinder due to all the holes in it. 
4) the shaft calculated as a slender rod.

So all that seems it would be easy enough.

Cheers

[TinselKoala]

When I think about it both methods are the same but just done a bit differently, so a valid result should be assured if it's done correctly and calculated without error. I'm still excited ! 

The way I see it for weighing the rotor parts I could do it like as follows, with reference to the links..

Link to the Engineering Toolbox "Flywheel Kinetic Energy"
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html

Flywheel shapes
http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html

1) the magnets can be calculated in total ( added together) as a hollow cylinder.
2) the squirrel cages can be calculated as solid cylinders
3) the plastic rotor plate can be calculated as a solid cylinder or maybe a hollow cylinder due to all the holes in it. 
4) the shaft calculated as a slender rod.

So all that seems it would be easy enough.

Cheers

Exactly. I forgot about engineeringtoolbox, that is a great site with lots of useful formulae. I referred to them a lot during the Mister Wayne days.

So since your rotor is symmetrical and made of regular shapes, it's straightforward, if not exactly "easy", to figure the MoI from measurements, weights and a good sketch or two.

It would be really interesting, I think, to compare the straight geometric method with the more empirical approach that MH suggested. I'll bet the answers will be really close. Validating measurement methods concurrently is always a good idea! We often take for granted that our tools are working properly but sometimes they aren't.

"Radian" measures always confused me until I realized that a "radian" is what you get when you take a line the length of the circle's radius, and wrap it around the circle's circumference. It takes 2 x pi or 6.28  "radiuses" to wrap all the way around the circle, so if you divide 360 degrees by 6.28 you get a bit over 57 degrees. It's the SI standard "derived" unit of the amount of rotation or angular measure, so if you use radians instead of degrees or RPM or something in the physics equations, your other SI units like Joules, Watts, seconds, kilograms and so forth all work out properly without needing extra constants of proportionality.