Pretty Simple Physics/Geometry Problem that Stumped Me

Question

Okay friends. So the setup here is that I was playing Silent Hunter III (World War II submarine simulator) and received a radio telling me the position and approximate velocity of a convoy I wanted to raid. I was trying to calculate the direction I should travel to intercept it most quickly. To solve this, I set up the problem as a 2D coordinate problem. I know that's not strictly speaking accurate with curvature of the earth, but I figured it was close enough and it still have me stumped.

So my position is $a_0$, the convoy I want to intercept has position $b_0$ and velocity $\dot b$. My submarine has a max speed (obviously) and $|\dot a| > | \dot b |$.

To make things a little bit easier on myself, I defined $b_0$ to be $(0,0)$. Also, to make this easier to conceptualize, I'll give you my actual figures I was working with, but I was trying to solve this algebraically to come up with a general solution.

$$a_0 = (-230, 75)\\ |\dot a| = 30 \\ b_0 = (0,0) \\ |\dot b| = 18.5$$

$b$ was traveling due southwest.

This was the work I managed to figure out:

$$ a(t) = \dot at+a_0 \\ b(t)=\dot bt\\ \text{The ships intersect when their position functions are equal.}\\ \therefore\dot at + a_0 - \dot b t=0\\ \text {Simplifying a bit:}\\(\dot a - \dot b)t+a_0=0$$

So I have two unknowns (i.e. $\dot a$ and $t$) but only one equation. I'm kind of stuck to figure out my second equation.

Edited to add: I should mention that my units are kilometers for distance and kilometers per hour for speed.

Use trig and the law of cosines. You have one side of a triangle $a$ to $b$ and the angle at vertex $b$ ($135 + \arctan \frac {75}{230}$) and you you have a ratio for the remaining two sides. — fleablood, Jul 02 '21 at 14:59
I thought about using the law of cosines, but I'm not sure how that would get me any further ahead. Like: $(b-a)^2 = (a't)^2 + (b't)^2 - 2 \dot a \dot bt^2 \cos \theta$. I have another equation, but also another unknown, to wit, theta. — Adam Gluntz, Jul 02 '21 at 15:05
Does this answer your question? A problem that gives rise to a differential equation — Display Name, Jul 02 '21 at 15:10
The differential equation answer is because the dog is stupid and always runs to where the rabbit is now. But in this case the game player is smart and is anticipating the enemy is moving and calculates it and moves in an efficient straight line. Different question altogether. — fleablood, Jul 02 '21 at 15:20
Here are two questions about the same problem: How to find direction of velocity V2 to reach an object travelling at velocity V1, such that it takes least time?, Intersection of two moving objects. There are probably others. — David K, Jul 02 '21 at 23:03

fleablood · Answer 1 · 2021-07-02T15:38:50.743

Let $c$ be the point of intersection.

You have a triangle $\triangle abc$ and the law of cosines says $(ab)^2 + (bc)^2-2(ab)(bc)\cos \angle abc = (ac)^2$.

$ab = \sqrt{230^2 + 75^2}$. $ac = |\dot a|t; bc=|\dot b|t$ (where $t$ is the time til intersection; an irrelevant variable which will cancel out) so $\frac {ac}{bc}=\frac {|\dot a|}{|\dot b|}$. And $m\angle abc = 135+\arctan(\frac {75}{230})$

(Actually you said the enemy is going due southwest but your image shows due southeast. I calculated for simplicity that $m\angle abc = m\angle (-230,75)(0,0)(-230,0) + m\angle(-230,0)(0,0)c$ and $m\angle (-230,75)(0,0)(-230,0)=\arctan \frac {75}{230}$ [ignoring orientation and negative values to keep it simple] and $m\angle (-230,0)(0,0)c = \text{due southeast} = 135$.)

So

$(230^2+75^2) + (bc)^2 - 2\sqrt{230^2 + 75^2}(bc)\cos (135+\arctan\frac {75}{230}) = (bc)^2( \frac {|\dot a|}{|\dot b|})^2$.

That's enough to solve $(bc)$ (there are two solutions but one of them will be a point where $b$ was in the past). And that's enough to give you $c$. ANd that gives you $\angle acb$.

Pretty Simple Physics/Geometry Problem that Stumped Me

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