Kind of esoteric, but..

[Jason]

Does anyone know offhand the formula to calculate the trajectory of an object (such as a torpedo) so that it impacts another object (such as a ship) moving in a known trajectory (constant) and speed(constant) the other object speed (torpedo) not being a constant but mathematically predictable (in other words a predictable variable in the equation)

I think it can be done as a multivariate equation, but I'm sure the right equation is much simpler.

[Jason]

No one? I stupidly put all my notes, references and textbooks in storage - I was tired of carrying them around Canada in my vehicle. :\

[Magicziggy]

Does a torpedo constantly assess the position of it's target and adjust course accordingly?

Or does it assess it's target's course and adjust course for predicted intercept.

Years ago, when doing my degree I wrote software for spacecraft rendevous and docking.
Not quite the same thing because the idea was to arrive a minimal closing velocity.

[Jason]

Does a torpedo constantly assess the position of it's target and adjust course accordingly?

Or does it assess it's target's course and adjust course for predicted intercept.

Years ago, when doing my degree I wrote software for spacecraft rendevous and docking.
Not quite the same thing because the idea was to arrive a minimal closing velocity.

In this case, the torpedo is 'dumb fire'. Once its trajectory is set it remains a constant. It should be simple, but I think I've gotten dumber in the past few months.

[Jason]

Oh I think I got it. I just need to find the gravitational acceleration in the given medium!

Sorry, I should have thought just a bit more.

[Magicziggy]

Oh I think I got it. I just need to find the gravitational acceleration in the given medium!

Sorry, I should have thought just a bit more.

I'm not familiar with torpedoes.

Are you modelling the motion of an object "fired" in the water?
In which case will it be slowing and falling?

Acceleration due to gravity is a constant on this planet. More or less.

[Jason]

Oh I think I got it. I just need to find the gravitational acceleration in the given medium!

Sorry, I should have thought just a bit more.

I'm not familiar with torpedoes.

Are you modelling the motion of an object "fired" in the water?
In which case will it be slowing and falling?

Acceleration due to gravity is a constant on this planet. More or less.

Yes, but I need to get the initial trajectory of the torpedo right.

No, it's a constant speed, the motion of the medium (water) will cause it to deviate slightly but it's statistically constant.

Oh.. isn't gravitational acceleration different depending upon the medium due to the specific densities involved?

There are a lot of variables, but I know the equation is simpler than the one I worked out. Not as accurate but close enough for about a 10km range.

[Jason]

It's sort of like if you wanted to launch a projectile at a piece of space junk ( an old satellite or whatever) and using any sort guidance after it was launched was not an option. Except I think that would involve using geodesics this is more straightforward and only uses linear vectors as it's such a short distance.

[Jason]

The formula has been used in naval warfare since at least WWI, I'm told it is fairly simple, officers who were not strictly 'mathematicians' were required to use it regularly.

Zilla are you out there?

[Magicziggy]

From wiki.http://en.wikipedia.org/wiki/Torpedo_Data_Computer

The effects of parallax and ballistics are minimal for small gyro angle launches because the course deviations they cause are usually small enough to be ignorable. U.S. submarines during World War II preferred to fire their torpedoes at small gyro angles because the TDC's fire control solutions were most accurate for small angles.[39]

The problem of computing the gyro angle setting is a trigonometry problem that is simplified by first considering the calculation of the deflection angle, which ignores torpedo ballistics and parallax.[40] For small gyro angles, θGyro ≈ θBearing - θDeflection. A direct application of the law of sines to Figure 3 produces Equation 1.

where

vTarget is the velocity of the target.
vTorpedo is the velocity of the torpedo.
θBow is the angle of the target ship bow relative to the periscope line of sight.
θDeflection is the angle of the torpedo course relative to the periscope line of sight.
Range plays no role in Equation 1, which is true as long as the three assumptions are met. In fact, Equation 1 is the same equation solved by the mechanical sights of steerable torpedo tubes used on surface ships during World War I and World War II. Torpedo launches from steerable torpedo tubes meet the three stated assumptions well. However, an accurate torpedo launch from a submarine requires parallax and torpedo ballistic corrections when gyro angles are large. These corrections require knowing range accurately. When the target range was not known, torpedo launches requiring large gyro angles were not recommended.[41]

Equation 1 is frequently modified to substitute track angle for deflection angle (track angle is defined in Figure 2, θTrack=θBow+θDeflection). This modification is illustrated with Equation 2.

where

θTrack is the angle between the target ship's course and the torpedo's course.

[Jason]

Ah! That's exactly what I was looking for! Thanks

The principle has surprisingly more uses than you'd think.

[Magicziggy]

Ah! That's exactly what I was looking for! Thanks

The principle has surprisingly more uses than you'd think.

I'm thinking "impregnation" right now. Those little suckers ain't guided.

[Jason]

Early space docking procedures always reminded me of something.

So.. now to begin constructing an nxn matrix circuit to finish the job..

[Gawdzilla Sama]

The formula has been used in naval warfare since at least WWI, I'm told it is fairly simple, officers who were not strictly 'mathematicians' were required to use it regularly.

Zilla are you out there?

Search Hyperwar, it's in there somewhere. Maybe in the PT boat section.