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]
That gives
$$\sqrt{1+ m^2}= \frac{mu}{v}$$
Squaring both sides,
$$1+ m^2= \frac{m^2u^2}{v^}$$
so
$$1= \left(\frac{u^2}{v^2}-1\right)m^2$$
$$= \frac{u^2- v^2}{v^2}m^2$$
$$m^2= \frac{v^2}{u^2- v^2}$$
$$m= \frac{v}{\sqrt{u^2- v^2}}$$
so in terms of the angle:
$$\theta= arctan(\frac{v}{\sqrt{u^2- v^2}}$$
You can see why the torpedo had better be faster than the tanker!

 

[tacman]

In order to calculate the angle to fire the torpedo, the u-boat captain would need to use the principles of projectile motion. This involves considering the initial velocity of the torpedo, the angle at which it is fired, and the gravitational force acting on the torpedo. By using these factors, the captain can calculate the trajectory of the torpedo and determine the angle at which it needs to be fired in order to hit the tanker.

First, the captain would need to determine the initial velocity of the torpedo. This would involve considering the speed of the torpedo and the speed of the u-boat. The captain would also need to take into account any external factors that could affect the velocity, such as wind or ocean currents.

Next, the captain would need to determine the angle at which the torpedo should be fired. This would involve considering the distance to the tanker and the speed and heading of the tanker. By using the distance and speed, the captain can calculate the time it will take for the torpedo to reach the tanker. By considering the heading of the tanker, the captain can determine the direction in which the torpedo should be fired.

Finally, the captain would need to take into account the gravitational force acting on the torpedo. This would involve considering the angle at which the torpedo is being fired and the acceleration due to gravity. By taking these factors into account, the captain can determine the exact angle at which the torpedo should be fired in order to hit the tanker.

Overall, calculating trajectories involves a combination of mathematical calculations and understanding of physical principles. By carefully considering all of the factors involved, the u-boat captain can determine the correct angle to fire the torpedo and increase their chances of successfully sinking the tanker.