Thread: Trig problem using bearings

Suppose that you are Torpedo Officer aboard the U.S.S. Skipjack. Your submarine is conducting torpedo practice off the coast of Florida. The target is 7200 m from you on a bearing of 276 degrees and is steaming on a course of 68 degrees. You have long-range torpedoes that will go 6400 m, and short-range torpedoes that will go 3200 m. Between what two bearings can you fire torpedoes that will reach the target's path if you use:
a. long-range torpedoes
b. short-range torpedoes


i have no idea what i am doing with this one.....lots of steps would be greatly appreciated

Originally Posted by jbrosrulez

Suppose that you are Torpedo Officer aboard the U.S.S. Skipjack. Your submarine is conducting torpedo practice off the coast of Florida. The target is 7200 m from you on a bearing of 276 degrees and is steaming on a course of 68 degrees. You have long-range torpedoes that will go 6400 m, and short-range torpedoes that will go 3200 m. Between what two bearings can you fire torpedoes that will reach the target's path if you use:
a. long-range torpedoes
b. short-range torpedoes


i have no idea what i am doing with this one.....lots of steps would be greatly appreciated

I do wish people who set you question that supposedly are abstracted
from the real world would at least know something about the context
in which they set their problems.

Anyone who knows anything about submarine weapons (present and historic)
will be cringing while reading this.

RonL
(Torpedo homing algorithm design consultant )

(When I get back to my other computer I will answer this if someone else
does not get to it first.)

Originally Posted by jbrosrulez

Suppose that you are Torpedo Officer aboard the U.S.S. Skipjack. Your submarine is conducting torpedo practice off the coast of Florida. The target is 7200 m from you on a bearing of 276 degrees and is steaming on a course of 68 degrees. You have long-range torpedoes that will go 6400 m, and short-range torpedoes that will go 3200 m. Between what two bearings can you fire torpedoes that will reach the target's path if you use:
a. long-range torpedoes
b. short-range torpedoes


i have no idea what i am doing with this one.....lots of steps would be greatly appreciated

The first thing you need is a diagram with relevant angles and distances marked. See attachment.

So we see that the closest the target will approch the Skipjack's position will be 7200 sin(28) ~= 3380.2 m

Now if the range to the point x is 6400m (so x is a limiting point for the use of long range weapons) Then the angle
between the normal from the Skipjack to the targets closest point of approach and the sight line to x is:

acos(3380.2/6400) ~= 58.1 degrees

So the required bearing is 276 + 62 - 58.1 = 279.9 degrees

and the other limiting angle is symmetricaly placed on the other side of the normal
so its bearing is:

276 + 62 + 58.1 = 396.1 = 16.1 degrees

There are no firing bearings for the short range torpedos, as the closest the target approaches the Skipjack is
~ 3380m which is beyond the range (3200m) of the short range weapons.

RonL

What did you draw your diagram with CaptainBlack?

Originally Posted by Jhevon

What did you draw your diagram with CaptainBlack?

With great skill

Real answer is PowerPoint.

RonL

Hello, jbrosrulez!

Oh-oh . . . I got different answers.

Code:

                          C
                          *
                      *  β  *
                  *           *
              *                 *
          *                       * 6400
    B *    28°                      *
              *                       *
                      *                 *
                 7200         *       θ   *
                                6°    *     *
                    - - - - - - - - - - - - - * A

We find that: .β .= .152° - θ

. . . . . . . . . . . sin β . . . sin28
Law of Sines: . ------- .= .------ . . → . . sin β .= .0.528155508
. . . . . . . . . . . 7200 . . . 6400

Then: .β .≈ .32°, 148°

Hence: .152° - θ .= .32°, 148° . . → . . θ .= .4°, 120°


The bearings are: .4° + 6° + 270° .= .280°
. .and: .120° + 6° + 270° .= .396° .= .36°

. . . or am I way way off?

Originally Posted by CaptainBlack

The first thing you need is a diagram with relevant angles and distances marked. See attachment.

So we see that the closest the target will approch the Skipjack's position will be 7200 sin(28) ~= 3380.2 m

Now if the range to the point x is 6400m (so x is a limiting point for the use of long range weapons) Then the angle
between the normal from the Skipjack to the targets closest point of approach and the sight line to x is:

acos(3380.2/6400) ~= 58.1 degrees

So the required bearing is 276 + 62 - 58.1 = 279.9 degrees

and the other limiting angle is symmetricaly placed on the other side of the normal
so its bearing is:

276 + 62 + 58.1 = 396.1 = 16.1 degrees

There is a typo here 396.1 is in fact the same thing as 36.1 degrees not
16.1 degrees

There are no firing bearings for the short range torpedos, as the closest the target approaches the Skipjack is
~ 3380m which is beyond the range (3200m) of the short range weapons.

RonL

Originally Posted by Soroban

Hello, jbrosrulez!

Oh-oh . . . I got different answers.

They are not different answers when I take 360 from 396.1 more carefully,
just of different precission (of course this was a deliberate error put in to
see who was awake ).

RonL

Originally Posted by CaptainBlack

(of course this was a deliberate error put in to see who was awake.)

But of course . . . I should have seen that.
Ya got me!