.......I have the updated version including the anti-gravity device at a new link.

John

I've split your posts out of the other topic as I think you have got lost in there.

Some questions/comments from the first part:

I think your equating N/Am with an acceleration is a weak start to your paper and likely to be the focus of any comments, better to bring it in later. You say the field must be an acceleration field, but a constant force will produce a constant acceleration, so it would be better to consider this as a force field. This would match better with the Tesla which has Newtons scaled by the Ampere meter. Teslas will not equal or represent an acceleration as the formulas cannot be resolved into m/s

^{2}, but you could introduce the N/Am after you have finished the example calculations and bring it in as a comparative/similarity.

In trying to calculate the value of g at the earth's surface I feel you have set yourself a difficult task. Although gravity is often described as a point source, in reality at the surface of the earth it is the sum of all the gravity vectors from the distributed mass of the earth, this is not a straightforward calculation and requires integration of all the sources. In particular not all the particles will give a downward component so a straight summation shouldn't give correct answer. Why not start with 2 lead balls and calculate the attraction between them. That way you will know the exact number of particles in the 2 balls and if the separation is significantly greater than the diameter of the balls you should be able to consider them as point sources.

I haven't had time to go through the rest of the maths yet, but will try to find time.