Analysis of Variance of Firing Angles and Hit Percentages on the Battlefield

[Chamberlain]

Or,

2 snipers from 400 yards, each one locked on a target !!



Chamberlain

[born2see]

Chamberlain wrote:

2 snipers from 400 yards, each one locked on a target !!

But then they'll really have to have this angle/gravity thing figured out.

B

[Chamberlain]

True B,

Then maybe we need 4 snipers, 2 for each one...1 to aim high, and 1 to aim low !

:woohoo:

Chamberlain

[Hancock the Superb]

Hancock The Superb wrote:

That makes absolutely no sense whatsoever. Since you are unwilling to do the calculation properly, I shall do it for you.

Without gravity:
d = Dist. to target: 100yd. (91.44M)
h = Height of target: 70.2in (1.78M)
angle = atan(h/d) = 1.12 degrees

With gravity:
speed of bullet: 950ft/sec (290M/S)
g = 9.8M/S**2
time of flight of bullet = t = 91.44/290 = 0.316 sec
drop of bullet during flight = hg = 1/2gt**2 = 0.49M
hr = real height of target = h - hg = 1.29M
angle = atan(hr/d) = 0.81 degrees

Gravity reduces the acceptance angle by 28% (good estimate Barrow)

If you have any desire to go into a field where quantitative measurements and calculations are required, I urge you, for your own sake, to listen to the criticism you receive and perform the measurement as they suggest. Only then can you determine its accuracy. No one gets particularly upset if you make a mistake in your assumptions or measurements. Science is a self correcting endeavor. However, people get very upset when you refuse to entertain the possibility of error.

If I am upsetting people, I must apologize. I am just trying to point out that my way works.

In addition, if I may be so bold to say, you and Barrow may have committed an error in assuming that the time of the bullet in flight is 0.316 seconds, or 91.4m divided by the muzzel velocity of 290m/s. In the physics I have learned, the muzzel velocity at an angle does not translate into horizontal velocity. According to my calculations below, it should actually be the 290m/s * cos(THETA).

The attachment Proof.jpg is no longer available

As you can see, the messy vertical displacement equation has a plethora of THETAs (due to the coorelation between the vertical initial velocity and the horizontal initial velocity to determine the time and speed), so I plugged it into my TI-84+ SE calculator and determined the THETA values of the intersection between the curve and the 0 meter vertical displacement line as well as the 1.78m vertical displacement line.

As you can see from the table on the right, the difference between the angles exactly follows my prediction of earlier (as seen by the chart on the first page): off only by the third significant figure on a couple of them.

Hopefully this moves us further towards the goal of coming up with an excellent equation simulating the ability of Civil War soldiers to hit a target.

I do appreciate all the constructive criticism I have recieved, it has definitely caused me to focus all aspects of physics involved as well as hopefully improving my communication skills.

As a side note, I find the calculated top angle for 50, 100, and 160 yards rather odd. However, I triple checked my work and my calculator inputs, and have decided that somewhere between 100 and 160 yards gravity plays a role such that the angle must increase in compensation to hit the top of the target.

[Barrow]

>As you can see from the table on the right, the difference between the angles
>exactly follows my prediction of earlier (as seen by the chart on the first
>page): off only by the third significant figure on a couple of them.

If I'm reading your table correctly, in all these calculations your target is
at a different elevation from your shooter, and your "distance" is the hypotenuse.
In your original post it seemed like your distance was the horizontal separation
between shooter and target, and they were both at the same elevation.

[Hancock the Superb]

>As you can see from the table on the right, the difference between the angles
>exactly follows my prediction of earlier (as seen by the chart on the first
>page): off only by the third significant figure on a couple of them.

If I'm reading your table correctly, in all these calculations your target is
at a different elevation from your shooter, and your "distance" is the hypotenuse.
In your original post it seemed like your distance was the horizontal separation
between shooter and target, and they were both at the same elevation.

I guess I don't quite understand what you mean. In all of my work, I have been firing at two points, the top of a person and the bottom, which are 1.78m apart vertically. You may have missed that in the flurry of posts.

[Marching Thru Georgia]

The gun is being shot from an elevation on 1.78M not 0 as you have drawn in your last post. There is no vertical component to the muzzle velocity. As I said before you are needless complicating the issue by trying to solve for 2 triangles. One is sufficient. Repeat your calculations using the method that I showed you. It is the correct way to solve the problem.

[Mayonaise]

I know nothing about physics that I didn't learn in highschool, but it seems to me you are solving two different problems, which is why you can't agree.

One is trying to determine that whatever angle hits the target, be it 30 degrees or 300, what range of angles up or down from that would hit the target of X height and Y range.

The other is trying to determine what angle would actually hit, which is why you keep talking about hitting the ground and other irrelevant facts.

Regardless, "march through georgia", even if you are correct you come across as arrogant and preachy. Go back and read how many times you condescendingly wrote "I am right you are wrong". That's the real lesson to learn in this thread.

[KG_Soldier]

And you, Mayonaise, come across as misspelled, white and fluffy, and quite tasty on a turkey and cheese sandwich.

[RebBugler]

:laugh:

And you, Mayonaise, come across as misspelled, white and fluffy, and quite tasty on a turkey and cheese sandwich.

:laugh: