# Mechanical Steel vs. Magnetic Levitation Composite Flywheels?

1. Nov 1, 2005

### BasketDaN

Spinning at their respective maximum velocities, approximately how much more energy will an ideal composite flywheel (magnetically levitated in a vaccum) be able to store than an ideal ball bearing steel alloy flywheel of the same size and shape? Thanks.

2. Nov 14, 2005

### BasketDaN

Anybody? ..

3. Nov 14, 2005

### Danger

Isn't it purely dependent upon the rotating mass (and distribution thereof) as opposed to what material is used?

4. Nov 14, 2005

### Staff: Mentor

I think he's asking about the difference that the support mechanism makes -- maglev versus bearings with physical contact. It seems like the maglev approach will still burn energy, because it will likely take some active feedback to keep the flywheel balanced while it spins, especially at high velocities. To the OP -- were you thinking of some passive maglev scheme to try to minimize the extra energy needed? Is passive maglev going to be sufficient? Do you have examples of maglev flywheels that you can point us to for reference?

5. Nov 14, 2005

### Danger

I see. I thought that the phrase 'ball bearing steel alloy flywheel' meant that the wheel was made out of bearing metal, not that it was supported by ball bearings.

6. Nov 14, 2005

### pervect

Staff Emeritus
Looks to be about 8x more energy / unit mass for composites from the figures at

http://www.aspes.ch/faq.html#Why%20composite%20materials [Broken]

(CFRP, some sort of carbon fiber composite, I guess, having the highest rating).

Last edited by a moderator: May 2, 2017
7. Nov 14, 2005

### BasketDaN

The difference only exists because composite flywheels are able to withstand far greater rotational velocities than are steel flywheel.s

Yeah, I found that site too,, do you think the specific strength is directly proportional to the maximum speed it can withstand?

8. Nov 15, 2005

### pervect

Staff Emeritus
Nope, velocity goes as the square-root of the specific strength, energy density goes directly as the specific strength.

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