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Weight change with rotation ? 
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#19
Apr412, 10:40 AM

P: 13

G is an imitation of g and his formula and value are very doubtful...



#20
Apr412, 11:48 AM

P: 365

I have corrected
2*m*G*M*H/100*(1/(R100)1/(R)) = 1964.52227397479936678121 compared to m*G*M/1*(1/(R+1)+1/R1/(R1)+1/R) = +3.08348993657429924912e6 m*G*M*H/100*(1/(R99)1/(R+1)+1/(R101)1/(R1)) = 1964.52227397494456785185 m*G*M/1*((1/(R99)1/(R100))+(1/(R101)1/(R100))) = 0.08363513764493721675e6 The sum is : 1964.522273974799366781213.08348993657429924912e61964.52227397494456785185+3.08363513764493721675e6 = 1e27 it's ok, the sum is 0. 


#21
Apr612, 05:34 AM

P: 365

I have thought about something I don't understand. When a screw is in water with bottom thread end with vertical surface. The torque on the bottom surface is canceled all along the thread with a differencial of pressure (up/down surface of the thread). With the pressure of water = f(h) it's ok, all torque are canceled, but now with pressure = f(h,Δh²) why the bottom torque is not greater than the differencial torque ? The difference is not big but must be to 0 I think ?
NB. We are for example on Earth with gravity = 9.81 m/s². The screw can only turn. 


#22
Apr612, 09:34 AM

P: 365

All is fine I understand the problem.
But for this thread I will calcultate if the torque is 0 : A circular thread (can only turn) in sandwich between 2 circular fixed thread. Up and down small film of water. See the drawing. The thread has only film of water up and down.Normally, this thread don't turn because the height of water is the same. But really, one side is more far from Earth (the up side). This give a torque ? 


#23
Apr612, 11:42 AM

P: 365

I drew the developpant of the helicoid (thread). This helicoid is in a sandwich between 2 fixed helicoids. Only film of water Up and Down surfaces.
The down torque = T1 = Radius_of_helicoid * Down_surface * density_of_water * height * Gt * Mt * ( g + (1/(R+250)  1/R ) ) The up torque = T2 = Radius_of_helicoid * Up_surface * density_of_water * height * Gt * Mt * ( g + (1/(R+350)  1/(R+100) ) ) With: g = 9.81 Mt = Mass of Earth Gt = Gravitational constant Up_surface = Down _surface k = Radius_of_helicoid * Down_surface * density_of_water * height * Gt * Mt T1 = k * ( g + (1/(R+250)  1/R ) ) = 9.8099999999938410347 * k T2 = k * ( g + (1/(R+350)  1/(R+100) ) = 9.80999999999384122804 * k There is a little difference ? 


#24
Apr612, 05:08 PM

P: 365

I find the problem, the force is perpendiculary to the surface and movement.
Another question is with a fixed column of water with a fixed mass under it. Like the mass is more at left, the pressure at left in the column is higher than at right. The water move inside the column ? Or another force balance this ? (No other gravity, only mass M and water). Mass of M is 100 or higher than mass of water for example, because there are forces in water alone, the up of water attrack the down, idem for the sides, maybe already a movement in water alone. 


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