Envelope of Trajectories - 2

The initial trajectory is $y \; =\; x\tan\theta - \tfrac{1}{80}x^2\sec^2\theta$ and, to begin with, we want to maximize $y$ as $\theta$ varies, for $0 \le x \le 15$ .

If $0 \le x \le 40\cot80^\circ$ , then $y$ is maximized for $\theta = 80^\circ$ , so the envelope has equation $Y(x) \; = \; x\tan80^\circ - \tfrac{1}{80}x^2\sec^280^\circ \hspace{2cm} 0 \le x \le 40\cot80^\circ$ If $40\cot80^\circ \le x \le 15$ , then $y$ is maximized when $\tan\theta = \frac{40}{x}$ , and hence $Y(x) \; = \; 20 - \tfrac{1}{80}x^2 \hspace{2cm} 40\cot80^\circ \le x \le 15$ Since the initial trajectory has gradient $\tan\theta - \tfrac38\sec^2\theta$ at $x=15$ , in the region $15 \le x \le 25$ the general trajectory is $y \; =\; 15\tan\theta - \tfrac{45}{16}\sec^2\theta + (x - 15)(\tan\theta - \tfrac38\sec^2\theta)$ and this is maximized when $\tan\theta = \frac{8x}{3(2x-15)}$ , and so the envelope has equation $Y(x) \; = \; \tfrac{125}{16} + \tfrac{7}{24}x + \frac{75}{2x-15} \hspace{2cm} 15 \le x \le 25$ Finally, in the region $x \ge 25$ , the general trajectory is $y \; = \; (x - 10)\tan\theta - \tfrac{1}{80}(x - 10)^2\sec^2\theta + 10(\tan\theta - \tfrac38\sec^2\theta)$ and this is maximized when $\tan\theta = \frac{40x}{x^2 - 20x + 400}$ , so the envelope has equation $Y(x) \; = \; \frac{20x^2}{x^2 - 20x + 400} - \tfrac{1}{80}(x^2 - 20x + 400) \hspace{2cm} x \ge 25$ We deduce that $Y(x) = 0$ when $x = 30 + 10\sqrt{5}$ , and so the complete envelope equation is $Y(x) \; =\; \left\{ \begin{array}{lll} x\tan80^\circ - \tfrac{1}{80}x^2\sec^280^\circ & \hspace{2cm} & 0 \le x \le 40\cot80^\circ \\[1ex] 20 - \tfrac{1}{80}x^2 & & 40\cot80^\circ \le x \le 15 \\[1ex] \tfrac{125}{16} + \tfrac{7}{24}x + \frac{75}{2x-15} & & 15 \le x \le 25 \\[1ex] \frac{20x^2}{x^2 - 20x + 400} - \tfrac{1}{80}(x^2 - 20x + 400) & & 25 \le x \le 30 + 10\sqrt{5} \end{array}\right.$ The area $A$ under this curve can be calculated explicitly, so $\begin{aligned} A & =\; \frac{25}{24} \left[\begin{array}{l}345 + 88 \sqrt{5} + 128 \sqrt{3} \tan^{-1}\big(\tfrac12\sqrt{3}\big) - 128 \sqrt{3} \tan^{-1}\left(\frac{2 + \sqrt{5}}{\sqrt{3}}\right) \\ + 36 \ln\tfrac{7}{3} + 192 \ln400 - 192\ln525 + 192\ln\big(3 + \sqrt{5}\big) - 256 \tan10^\circ \end{array}\right] \\[2ex] & = \; \frac{25}{24}\left[345 + 88\sqrt{5} - 128\sqrt{3}\tan^{-1}\left(\tfrac{1}{33}(7\sqrt{15} -6\sqrt{3})\right) + 36\ln\tfrac73 + 192\ln\left(\tfrac{16}{21}(3 + \sqrt{5})\right) - 256\tan10^\circ\right] \\ & \approx \; 717.5331561 \end{aligned}$ and hence $\lfloor A \rfloor = \boxed{717}$ .

This was a fun one. Congrats on the 100 problems. I did a triple sweep, so the computational complexity was huge.

1) Sweep a "measurement x value" $(x_m)$ over a range in discrete steps.

2) For each $x_m$ , sweep the launch angle $\theta$ over the allowable range in discrete steps.

3) For each $\theta$ , run a time simulation to see the $y$ value when $x = x_m$ . If $x$ never reaches $x_m$ , that trajectory's $y$ value counts as zero. Store the maximum $y$ value $(y_{max})$ for each $x_m$ .

4) The incremental area $dA = y_{max} \, d x_m$

In my highest resolution sweep, $d x_m = 0.1$ , $d \theta = 0.1 ^\circ$ , and $dt = 10^{-4}$

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import math

m = 1.0
u = 20.0
g = 10.0

deg = math.pi/180.0

####################################

dxm = 1.0
dtheta = 1.0*deg
dt = 10.0**(-4.0)

dxm = dxm/10.0
dtheta = dtheta/10.0
dt = dt/1.0

####################################

A = 0.0

xm = 0.0

while xm <= 55.0:

    ymax = 0.0

    theta = 0.0

    while theta <= 80.0*deg:

        t = 0.0

        x = 0.0
        y = 0.0

        xd = u*math.cos(theta)
        yd = u*math.sin(theta)

        xdd = 0.0
        ydd = -g

        while (x <= xm) and (y >= 0.0):

            x = x + xd*dt
            y = y + yd*dt

            xd = xd + xdd*dt
            yd = yd + ydd*dt

            if (x >= 15.0) and (x <= 25.0):
                F = 10.0
            else:
                F = 0.0

            Fx = 0.0
            Fy = F - m*g

            xdd = Fx/m
            ydd = Fy/m

            t = t + dt

        if y > ymax:
            ymax = y

        theta = theta + dtheta

    dA = ymax*dxm
    A = A + dA

    print xm,ymax

    xm = xm + dxm

####################################

print ""
print ""
print dxm
print dtheta
print dt
print ""
print A

########################

#1.0
#0.0174532925199
#0.0001

#712.168927954

########################

#0.5
#0.00872664625997
#5e-05

#717.386333522


########################

#0.2
#0.00349065850399
#5e-05

#716.562647831

########################


#0.1
#0.00174532925199
#0.0001

#717.565745138