How Would a U-Boat Captain Calculate the Angle to Fire a Torpedo at a Tanker?

[george01]

Say in World War II a u-boat captain is trying to sink a tanker. He knows the distance to the tanker, its speed and heading (which are assumed to be constant) and the speed of the u-boats torpedos. How would he calculate what angle to fire the torpedo to hit the tanker?? (Assuming the u-boat firing calculator thing was broken and everyone else on board was busy)

 

[HallsofIvy]

Sounds pretty straight forward to me. Let's assume you set up a coordinate system so that the u-boat is at (0,0) and, at t=0, the tanker is on the x-axis moving parallel to the positive y-axis. If its speed is v, and distance is d, then its position at time t is (d, vt). Since the torpedo's path will be a straight line passing through (0,0), it can be written y= mx where m is the slope (tangent of the angle). But we need that in terms of t: Okay, if x= at, then y= mat and the distance along the line from the u-boat (0,0) to the torpedo is (by Pythagorean theorem) $\sqrt{a^2t^2+ m^2a^2t^2}= at\sqrt{1+ m^2}$. If the torpedo's speed is u, then we have $ut= at\sqrt{1+ m^2}$ so
$$a= \frac{u}{\sqrt{1+ m^2}}$$.

The torpedo will cover the horizontal (x-direction) distance d when
$$at= \frac{ut}{\sqrt{1+ m^2}}= d$$
In otherwords, when
$$t= \frac{d\sqrt{1+ m^2}}{u}$$
The y- component at that time, y= mx= md must be equal to the distance the tanker has traveled,
[tex]vt= \frac{vd\sqrt{1+ m^2}}{u}= md[/itex]<br /> That gives<br /> $$\sqrt{1+ m^2}= \frac{mu}{v}$$<br /> Squaring both sides,<br /> $$1+ m^2= \frac{m^2u^2}{v^}$$ <br /> so <br /> $$1= \left(\frac{u^2}{v^2}-1\right)m^2$$<br /> $$= \frac{u^2- v^2}{v^2}m^2$$<br /> $$m^2= \frac{v^2}{u^2- v^2}$$<br /> $$m= \frac{v}{\sqrt{u^2- v^2}}$$<br /> so in terms of the angle:<br /> $$\theta= arctan(\frac{v}{\sqrt{u^2- v^2}}$$<br /> You can see why the torpedo had better be faster than the tanker![/tex]