Mostly Permanent Magnet Motor with minimal Input Power

[Magluvin]

Now, whether it be N or S pole fields detected by the sensor, the lines of force are measured/detected  by going through the face of the sensor, the larger flat sides. If the field lines go through the edges, the sides of the sensor, top to bottom or side to side, there will be little to nothing detected or measured.

So below shows the lines of force around the coil leg with the sensor orientation as you have demonstrated. Notice how the lines of force do go through the face of the sensor when at the left or right, but in the middle, the lines go through the top to bottom of the sensor. Lets call the end of the sensor with the 3 leads the bottom of the sensor.

More next post

Mags

[Magluvin]

Here we have the sensor in the orientation that I suggested earlier.

Notice how the lines of force with the sensor in the middle, the sensor should read more than to the left or right, sorta opposite of the way you measured it in the vid. Not that you did anything wrong. Im just giving a view of looking at the fields in a second dimension to show a more complete view of the fields. A third dimension of the field would be to face the sensor with the length of the wire/leg of the coil.

Mags

[Khwartz]

Ron, please do keep up your experiment and share your findings as it is worth more then words.

Khwartz, can you please calculate how much Joule energy it takes at Unity to lift 2.35Kg. 1mm, 2mm and 3mm in 1 second.

Thanks for your time and help

Luc

Yep! Except that if "by second" it means you ask for Watts and not Joules ^_^ but I will give you both:


□ W [J] = M [kg] × g [m.s^-2] × h [m]

□ P [W] = W [J] / T [ s]


● For 1 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 1/1000 [m]

= 23.5/1000 [kg.m^2.s^-2] = 0.0235 [J]


P [W] = 0.0235 [J] / 1 [ s] = 0.0235 [W]


● For 2 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 2/1000 [m]

= 47/1000 [kg.m^2.s^-2] = 0.047 [J]


P [W] = 0.047 [J] / 1 [ s] = 0.047 [W]


● For 3 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 3/1000 [m]

= 70.5/1000 [kg.m^2.s^-2] = 0.0705 [J]


P [W] = 0.0705 [J] / 1 [ s] = 0.0705 [W]


So the potential gravitational energy of the mass is increased proportionally to the height and the power in Watt goes with the energy in Joules, in term of value (but not the same nature of quantity).

If not clear enough, please just tell me

[gotoluc]

Thanks Khwartz for the calculations

Can you please also go back the the previous page and delete the oversize pic you posted.
It is causing problems for the page to load.
Also, I worked all day on my guide system and it's now 90% done, so I'm not going to change anything unless mine fails

Thanks for your help

Luc

[Khwartz]

You're most welcome for the calculations 

The big picture is already deleted.

Nice that you are no more worried about your guiding system to stick the outside magnet between the your steel plates and that you have well advance on