# Simple Harmonic Motion and Battleships

1. Nov 6, 2007

### Irelandp

Hi,

Recently, a friend and I were watching a 2nd world war film where a battleship was shelling land based fortifications. The discussion arose on how the ship (which was moving up and down due to the waves in the sea) could accurately target the fortifications.

I remember being told that the gunners waited until the ship was at the same state in the cycle before firing. This lead to duscussing what would be the best part of the cycle to fire. My view was that either the maximum or minimum of the cycle would be best, my friend thought the middle of the cycle would be better, as this give twice the number of opportunities to fire. I feel the only way to resolve this would be to calculate the positional probability distribution function for the gun on the ship undergoing simple harmonic motion.

I would appreciate help with this. Namely

Assuming the battleship is moving up and down on the waves where its vertical displacement about a mean is given by:

d = A sin (wt)

At any given arbitrary time T what would be the probability of finding the ship at a given displacement from the mean D.

Thanks

2. Nov 6, 2007

### Integral

Staff Emeritus
I believe (could be wrong) that WWII era ships had gyro stabilization for gun turrets, thus the gun crews did not have to worry about when to fire, the fire control systems did the work of holding the gun still for them.

Drop back a few years to a time when the gun crews made the call then it would make sense to fire at the extremes of motion, this is when the vertical velocity is a minimum.

3. Nov 6, 2007

### Irelandp

Your point about the extremes having the minimum vertical velocity, is central to my argument. However, the tennant of my friend's argument is that by reference to a distance point on land, using a telescope fixed to the ship, the accuracy of the timing when you fire is not a problem, therefore you are better choosing the mid point of the cycle as it gives you twice as many shots.

My counter is that my gut feeling suggests the time spent at the extremes is greater than that spent at the mid points, unfortunately, I am not able to prove it!

4. Nov 6, 2007

### Integral

Staff Emeritus
In order to hit something you must have good knowledge of the initial velocity of the projectile. Firing when the vertical motion is at a maximum increases the error in your knowledge of the IV. You also have 2 extremes of motion in every cycle, not sure how he gets more opportunities to fire???

But again, gyro controlled turrets makes this entire argument academic at best.

I am not sure why you should have trouble proving that you spend less time at mid motion, the velocity is greatest there... done.

Last edited: Nov 6, 2007
5. Nov 6, 2007

### clem

In Lord Nelson's fleet, they timed the firing with the roll of their ship to determine whether they would fire high at the rigging (or for more range) or low at the hull.

6. Nov 7, 2007

### phlegmy

as well as gyros for the gun, i believe some modern war ships have submerged "fins" which act to stop the boat rolling, or return it to level as quick as possible.

7. Nov 8, 2007

### rcgldr

An article referreing to gyro usage dating back to at least 1938:

http://www.navweaps.com/index_tech/tech-074.htm

8. Feb 4, 2008

### venkatmn

the probability density function......

Dear Irelandp,

Actually, the general equation for a simple harmonic function would be of the form

x=a.sin(wt+φ) (1)

where 'a' is the amplitude and 'φ' is the initial phase and 'w' is the angular frequency of the motion. Among these three variables a and w are predetermined because the gunners would presumably know the extrema distances and the speed of the swing. What they dont know is the inital phase 'φ'. The initial phase 'φ' has to be distributed uniformly between any period of '2*pi', or in other words, the probability density function of phi is

f(φ)=1/(2*pi) -pi<φ<pi

Having known f(φ), now the problem reduces to that of finding the pdf of ‘x’. We can recast Eq.(1) as

φ=arcsin(x/a)- ωt

Differentiating the above equation gives us

dφ/dx=1/√(a^2-x^2),

also Eq.(1) has two solutions in the interval (-π, π) for any ‘ωt’. Applying the formula for functions of a one random variable we get(Refer Papoulis).

f(x)= 1/(2π√(a^2-x^2))+ 1/(2π√(a^2-x^2))= 1/(π√(a^2-x^2))
|x|≤ a (2).

As we can see from Eq.(2) the probability diverges for values of ‘x’ as it approaches ±a.
So the gunners will have a stable ground only at the extremes( maximum or minimum)

Hope this helps.

Regards,
Venkatmn

9. Feb 4, 2008

### Irelandp

Thanks this is eactly what I was looking for.