Thane Heins Perepiteia.

[innovation_station]

 

good work guys keep it up


is

[polarbreeze]

THE PRIMARY HAS A PEAK MAGNETIC FLUX TRANSFER WINDOW WHICH IS DICTATED (RIGHT NOW) BY THE NUMBER OF TURNS OF WIRE ON IT. FOR EXAMPLE LUC INCREASED OUR PRIMARY YESTERDAY BY 15 TURNS AND OUR EFFICIENCY INCREASED BY 20%.

Sounds like the extra 15 turns bring you closer to a good impedance match. What you should find is that as you change the number of turns, you'll eventually get to an optimum where the impedance is matched as close as it can be, then adding more turns will make the efficiency drop off again. A good way to present the results might be as a graph with # of turns on the X axis and efficiency on the Y axis.

Some cautions: (a) you need to be very careful about your power measurements because your voltage and your current are not in phase - you can't just multiply them together to calculate the power. And (b), I suggest you revisit your method of calculating power factor because with the voltage and current not in phase the power factor can't be unity.

[polarbreeze]

Test # 4  -  1 K ohm LOAD
Primary I/P   
1.75 volts
1.88 amps
PF = 1
Input Power = 3.29 watts (17.8 % decrease)
Output Power = 1.05 watts @ 32.4 v (1887.7 % increase)

Thane/Luc, there's a methodology issue that I think you need to take into account. Here is an example from the earlier batch of tests but it will apply also to any present and future test. The issue is that you're not taking account of the extra power dissipated in the variac (in test #4 in this case) when you let the input voltage drop, details below. Caution: as I said before, these numbers will not be exactly right because the voltage and the current are not exactly in phase; however, it serves as an approximation to give an understanding of where the missing power went:

Between your tests 3 and 4, the increased load introduced a voltage drop in the variac of 0.25V (2.0V dropped down to 1.75V). Since the current is now 1.88A there is an additional 0.47W (0.25 x 1.88) being dissipated in the variac that wasn't being dissipated before. Because you're not keeping the input voltage constant at 2V, what you're really doing is causing some of the power that was previously dissipated in the coils to be dissipated in the variac instead. But it's being dissipated either way so it does need to be counted. If you don't count that power in your equations you're missing a key piece of where the input power went. To compare apples to apples, you need to consider the input power in test 4 to be 3.76W (ie 3.29 + 0.47). This is actually MORE than in test 3, which was 3.6W. The true input power did not go down: it went up.

As a corollary to this, I think a much more representative way to conduct the tests would be to ensure that the input voltage to the primary is always the same, adjusting the variac at each test if necessary to achieve that. This will ensure that the power dissipated in the variac is automatically eliminated from all the comparisons. Otherwise you may find that people will be able to challenge the results as not being apples-for-apples comparisons.

[gotoluc]

Hi All,

please find a link to megaupload to a single zip file of 4 pictures of the aluminum dune buggy that the Ottawa University has donated to us.

Download link: http://www.megaupload.com/?d=5KLKQU8R

We will be removing the gasoline engine and installing Thane' s accelerating charging electric motor.

Stay tuned

Luc

[JustMe]

Thanks...  Luc's pictures in the .rar file are large, high res images of very nice quality and well worth downloading.  But again, for the more casual user who may not be equipped to easily deal with the archive format I offer them here as smaller, lower quality images.  Note to JackS: There are FOUR (4) (IV).