The thorny way of truth - Free Energy Community

- 242 - MarinovADDENDUMSee the paper "Propulsive and..published in TWT-IV, p. 136.The anonymous referee of the above paper who is supporter of Ampere's formula for thInteraction of two current elements and who considers Grassmann's formula (1) as wrong ,presented the following objection:The Ampere law easily explains the various motoring actions shown in this paper.The propulsion forces are reaction forces between parts of the circuit which canmove relative to each other. Consider figure 1. When ABC and FGH are stationaryand sliding contacts are provided at C and F, the bridge will move in the f^-direction, but the f^- forces are not acting on DE. They are longitudinal forces in CDand FE produced by repulsion across the corners from BC and CF.The same longitudinal forces also explain the rotation of the motor when BC andGF are metal disks with sliding contacts to CD and FE.The Ampere force predicts that Marinov's motor will not rotate if C and F arewelded conductor junctions and AB and GH are made very long compared to DE. Thecurrent from a battery or other source would then have to be introduced with slidingcontacts at A and H. Also the battery leads would have to be kept well awayfrom the motor. If the motor still rotates with these modifications, then I acceptMarinov has proved his point of selfpropulsion.I agree that Ampere's formula predicts for my rotating Ampere bridge a torque in thesame direction as predicted by Grassmann's formula.However the referee is not right when asserting that if the wire BCDEFG in fig. 1 issolid with sliding contacts at the points B and G and with a rotational degree of freedomabout the axis ABGH, then according to Grassmann's formula a rotational moment(torque) should exist (according to Ampere's formula, of course, the torque in such acase must be null ).Now I shall show that also according to Grassmann's formula the rotational momentmust benull.Indeed, if assuming BC = CD = DF = EF = FG = a and if transferring the origin of thereference frame from point D to point B, the moment of the forces f-. which will drivethe bridge clockwise about the axis ABGH, will be given by the following formula (seeformula (6))Pi, .2R(-2)x/df3y = (u^I^/27i)R/(dx/x)x. (18)^0^0