# [Maxima] Shouryya Ray's equation

Henry Baker hbaker1 at pipeline.com
Thu May 31 08:58:57 CDT 2012

```Thanks very much for this posting.  It sounds very interesting.

Do you have a reference or link to this teenager's paper?

At 06:24 AM 5/31/2012, Jaime Villate wrote:
>Hi,
>you might have seen a picture of the Indian teenager holding a poster with the equation he recently
>discovered. If you are curious to know what it is about, here is a possible derivation of the equation.
>
>The motion of a projectile on a vertical plane xy is due to two external forces, its weight [0, -m*g],
>and the air resistance (Fr) which is proportional to the square of the speed v (v=sqrt(vx^2+vy^2) and
>it is always opposite to the velocity [vx,vy]
>
>   Fr/m = -c*v*[vx,vy]
>
>where m is the mass and c is a constant which depends on geometric shape and size, and the air density.
>We will also use the slope of the trajectory, s=tan(vy/vx).
>
>Let us first declare the dependence of vx, vy and s on time t
>
>(%i2) depends([vx,vy,s],t)\$
>
>The total force divided by the mass gives the acceleration vector and thus we have two differential
>equations for vx and vy
>
>(%o2) [vx(t),vy(t),s(t)]
>(%i3) eq1: diff(vx,t) = -c*vx*sqrt(vx^2+vy^2);
>
>(%o3) 'diff(vx,t,1) = -c*vx*sqrt(vy^2+vx^2)
>(%i4) eq2: diff(vy,t) = -c*vy*sqrt(vx^2+vy^2)-g;
>
>replacing vy by s in the equations, they become
>
>(%i5) eq3: ev(eq1,vy=s*vx,ratsimp);
>
>(%o5) 'diff(vx,t,1) = -c*sqrt(s^2+1)*vx*abs(vx)
>
>(%i6) eq4: ev(eq2,vy=s*vx,diff,ratsimp);
>
>(%o6) s*'diff(vx,t,1)+'diff(s,t,1)*vx = -c*s*sqrt(s^2+1)*vx*abs(vx)
>
>and we can write them in a simpler form
>
>(%i7) sol: first (solve([eq3,eq4],[diff(vx,t),diff(s,t)]));
>
>(%o7) ['diff(vx,t,1) = -c*sqrt(s^2+1)*vx*abs(vx),
>       'diff(s,t,1) = -g/vx]
>
>The only horizontal force is the component of the air resistance, which is proportional to v^2 and,
>therefore, vx will either be positive or negative, but not both. Let us assume that the projectile
>was launched in the x-positive direction
>
>(%i8) assume(vx>=0)\$
>
>We can know now obtain an ordinary differential equation (ODE)  for s and vx
>
>(%i9) eq5: 'diff(s,vx) = ratsimp(rhs(sol[2])/rhs(eq3));
>
>(%o9) 'diff(s,vx,1) = g/(c*sqrt(s^2+1)*vx^3)
>
>which is a very simple separable ODE and its solution is
>
>(%i10) eq6: ode2(eq5,s,vx);
>
>(%o10) (c*asinh(s)+c*s*sqrt(s^2+1))/(2*g) = %c-1/(2*vx^2)
>(%i11) eq7: expand(first(solve(eq6,%c)));
>
>(%o11) %c = 1/(2*vx^2)+c*asinh(s)/(2*g)+c*s*sqrt(s^2+1)/(2*g)
>(%i12) eq8: subst(sqrt(s^2+1)=sqrt(vx^2+vy^2)/vx,eq7);
>
>(%o12) %c = c*s*sqrt(vy^2+vx^2)/(2*g*vx)
>          +1/(2*vx^2)+c*asinh(s)/(2*g)
>(%i13) eq9: subst(s=vy/vx,eq8);
>
>(%o13) %c = c*asinh(vy/vx)/(2*g)+c*vy*sqrt(vy^2+vx^2)/(2*g*vx^2)+1/(2*vx^2)
>
>this last result is the famous equation. If the initial speed was v0 with an angle a0 above the horizontal,
>we have
>
>(%i14) ic1(eq9,vx=v0*cos(a0),s=tan(a0));
>
>(%o14) (cos(a0)^2*c*v0^2*asinh(vy/(cos(a0)*v0))
> +c*vy*sqrt(vy^2+cos(a0)^2*v0^2)+g) /(2*cos(a0)^2*g*v0^2)
>  = (c*vx^2*asinh(vy/vx)+c*vy*sqrt(vy^2+vx^2)+g)/(2*g*vx^2)
>
>Regards,
>Jaime

```

More information about the Maxima mailing list