1.0  Creating a kind of artificial gravity

Just an idea

 

Should it be possible to create a sort of artificial gravitity/force? (for flying cars, skateboards, helicopters without roters, propelling spacecrafts ..)

 

Lets consider a  rod  turning with a constant angular velocity, and a mass connected to this rod, that can be moved alongside the rod.

 

In position 1 the mass is at its greatest distance r1 from the axis of the rod.

  

 

 

For example: m = 0,5 kg;  r1 = 0,20 m;  T =  0,1 s  (10 revolutions/sec)    -> F1394,8 N

 

 

Now the rod has turned through θ = 180   and the mass has been moved towards the centre.

 

For example: m = 0,5 kg;  r2 = 0,05 m;  T =  0,1 s  (10 revolutions/sec)    -> F2 98,7 N

 

F1 - F2  = 296,1 N (= about 30 kgf)

 

 

The rod has turned through an angle  θ2 from position 1 to position 2 with the mass m still at its greatest distance r1 from the axis.

A mechanism connected at the rod (for example a kind of electromagnet) starts pulling the mass m towards the centre.

 

In figure 6 a possible path of mass m is indicated (perhaps some points will be in reality a little bit more smoothed, but I intuit that this is not so important).

If we express the path of the mass m as a function of time, we should be able to calculate the force (as a function of time) that has to be exerted on the mass m to oblige it to follow this path.

This force will be equal, but opposite in sign to the force that the mass m exerts on the rod (on the axis of the rod)

 .

θ(t) = 360/(2π) . ω.t  (in degrees; ω = 2π/T ; T = time of one revolution; the angular velocity is constant)

Programming in Pascal/Delphi:

var m, r1,r2, T: real;
te2, te3, te5, te6 : integer; {angles teta}
mm: Tm_array;

Procedure Calculate;
var radius, radius0, sumax, sumay, tt,tt0, w,
dxdt, dydt, dxdt2, dydt2, x,y, Fx, Fy,
dxdt0, dydt0, dxdt20, dydt20, x0,y0, Fx0, Fy0,
teh, teh0, dtradius, dt2radius: real;
integraalFydt, integraalFxdt : real;
Fymean, Fxmean: real;
dd, dt: real;
te, q, cc: integer; {angle teta=te and angle fhi=q}
strh: string;

begin
T:=0.1; {10 rotations per sec}
w:=2*pi/T; {= 628,3 rad/s angular velocity in rad/sec}
m:=0.1; {0,1 kg}
r1:=0.2; {20 cm}
r2:=0.05; {5 cm}
te2:=50; {50 degrees}
te3:=90;
te5:=230;
te6:=270;
cc:=10; {to get a more accurate calculation}

sumax:=0; {for every degree the acceleration ax of m in the x-direction is calculated; the program sums all the ax}
sumay:=0; {idem for the y-direction}

For te:=1 to (te2*cc) do {we choose that for te=80 = 0,44.pi the mass m starts moving towards the axis: point 2 in fig.6; 1 degree=360/T =360 w/2pi sec)}
Begin
teh:=te/cc; {angle teta} {if te=100 then teh=1 degree, if te=8000 then teh=80 degree, with cc=100}
tt:= 2*pi* teh/(360*w); {time} {if teh=360 then tt=T, angular velocity = constant, so tt = constant * teh}
x:= r1*sin(w*tt); {for example, te=7000 -> angle =70, tt= }
y:= r1*cos(w*tt);
mm[te].x:=x;
mm[te].y:=y;

dxdt:= w*r1*cos(w*tt); {speed}
dydt:= -w*r1*sin(w*tt);
mm[te].vx:=dxdt;
mm[te].vy:=dydt;

dxdt2:= -w*w*r1*sin(w*tt); {accelaration}
dydt2:= -w*w*r1*cos(w*tt);

sumax:=sumax+dxdt2; {dt=T/(360*cc} {after each dt the ax and ay is calculated and all ax and ay are summed together}
sumay:=sumay+dydt2;
End;

For te:=(te2*cc+1) to (te3*cc) do {te2=50, te3=90}
Begin
dd:=te3-te2; {40}
teh:=te/cc; {if te=8001 then teh:=80,01 degree, with cc=100}
tt:= 2*pi* teh/(360 *w);
radius:=(r2+ (r1-r2)*( (te3- teh)/ dd ) ); {between 50 and 90 the radius decreases proportionally with the angle (and so with the time) }
dt:= T/(360*cc);
x:= sin(w*tt)*radius;
y:= cos(w*tt)*radius;
mm[te].x:=x;
mm[te].y:=y;
dxdt:=(mm[te].x-mm[te-1].x)/dt;

dtradius:= (-r1/dd+r2/dd)*360/T; {(-0.1*r1+0.1*r2)*360/T;} {de 1e derivative}
dt2radius:= 0; {the 2e derivative}



{dxdt:= w*cos(w*tt)*radius + sin(w*tt)*dtradius;
dydt:= -w*sin(w*tt)*radius + cos(w*tt)*dtradius;}

dxdt2:= -w*w*sin(w*tt)*radius + w*cos(w*tt)*dtradius + w*cos(w*tt)*dtradius+0;

dydt2:= - w*w*cos(w*tt)*radius - w*sin(w*tt)*dtradius - w*sin(w*tt)*dtradius+0;

{sumax:=sumax+dxdt2;
sumay:=sumay+dydt2;}
End;

For te:=(te3*cc+1) to (te5*cc) do {te5=230}
Begin

teh:=te/cc;
tt:= 2*pi* teh/(360 *w) ; {if teh =180 then tt:= 0,05 is T/2}
x:=r2*sin(w*tt);
y:=r2*cos(w*tt);

dxdt:= w*r2*cos(w*tt);
dydt:= -w*r2*sin(w*tt);

dxdt2:= -w*w*r2*sin(w*tt);
dydt2:= -w*w*r2*cos(w*tt);

sumax:=sumax+dxdt2;
sumay:=sumay+dydt2;
End;

For te:=(te5*cc+1) to (te6*cc) do {te6=270}
Begin
dd:=te6-te5; {40}
teh:=te/cc;
teh0:=(te-1)/cc;
tt:= 2*pi* teh/(360 *w);
tt0:=2*pi* teh0/(360 *w);
radius:= (r1 - (r1-r2) * ((te6 - teh)/dd) ); {between 230 and 270 the radius increases proportionally with the angle and so with the time}
radius0:= (r1 - (r1-r2) * ((te6 - teh0)/dd) );
dtradius:= (r1/dd - r2/dd) *360/T; { (0.1*r1-0.1*r2)*360/T;}

x:= sin(w*tt) * radius;
y:= cos(w*tt) * radius;
x0:= sin(w*tt0) * radius0 ;
y0:= cos(w*tt0) * radius0;

dxdt:= (x-x0)/(tt-tt0); {w*cos(w*tt)*radius+sin(w*tt)*dtradius}
dydt:= (y-y0)/ (tt-tt0); {-w*sin (w*tt)*radius+cos(w*tt)*dtradius;}

dxdt2:= -w*w*sin(w*tt)*radius + w*cos(w*tt)*dtradius + w*cos(w*tt)*dtradius + 0;
dydt2:= -w*w*cos(w*tt)*radius - w*sin(w*tt)*dtradius + - w*sin(w*tt)*dtradius + 0;

{ sumax:=sumax+dxdt2;
sumay:=sumay+dydt2; }

End;

For te:=(te6*cc+1) to (360*cc-1) do
Begin
teh:=te/cc;
tt:= 2*pi* teh/(360 *w) ;
x:=r1*sin(w*tt);
y:=r1*cos(w*tt);


dxdt:= w*r1*cos(w*tt);
dydt:= -w*r1*sin(w*tt);

dxdt2:= -w*w*r1*sin(w*tt);
dydt2:= -w*w*r1*cos (w*tt);

sumax:=sumax+dxdt2;
sumay:=sumay+dydt2;
End;
{suamay = cc*360 }


Fy:=m*sumay; {= m * (dFy1+dFy2+dFy3+....) }
Fx:=m*sumax;

integraalFydt:=Fy*(T/(360*cc)); {= dFy1*dt+dFy2*dt+... = (dFy1+dFy2+..)*dt = Fmean * T} {dt=T/(360*cc}
integraalFxdt:=Fx*(T/(360*cc));

Fymean:=integraalFydt/T;
Fxmean:=integraalFxdt/T;

str(Fymean:0,strh);
form1.label2.caption:=strh;

str(Fxmean:0,strh);
form1.label3.caption:=strh;

end;

procedure TForm1.Button1Click(Sender: TObject);
begin
Calculate;
end;

end.

 

Well get a net accelarion upwards?

Lets run the program and find out ... ??????

The result are strange values..   probably cause: when the mass starts to move towards the centre, the speed alongside the radius has suddenly a value, which in reality is not possible. It should feel an accelaration towards the centre and increase the speed towards the centre as a function of time, and later on get a deaccelaration and then slowly approaching the centre.

----------------

Lets calculate the following:  (should be quite easy to contruct..)

 

The mean accelaration in the y-direction is the integral of f"(t) dt over T = > the surface of ff" under the horizontal axis = the surface of ff" above the horizontal axis.

f(t) is the primitive function of f"t), if you follow the red curve of f(t) you will see easily that the surface under = surface above over one period T.

=> the mean accelaration in the y-direction = zero

Newton's first law holds ...


 

https://www.derivative-calculator.net/

https://patents.google.com/patent/US3683707A/en  (a device to convert a circular movement into a linear force)

https://www.youtube.com/watch?v=r7NPx3dRpUw

https://www.youtube.com/watch?v=obnxoJsBDGQ

 

 

 

 

 

14 February 2020      by  Rinze Joustra        www.valgetal.com