Author Topic: Some things I won't miss about AH1 (Read 1658 times)

Eagler · « **Reply #15 on:** May 29, 2004, 09:41:26 AM »

Quote

Originally posted by Eagler
AH1 hit bubble = O
AH2 hit bubble = .

Originally posted by SUPERFLY
AH doesn't have a hit bubble.

ok then:

AH1 Bullet Size = = O

AH2 Bullet Size = = .

beet1e · « **Reply #16 on:** May 29, 2004, 11:07:27 AM »

LOL Eagler!

Blammo · « **Reply #17 on:** May 29, 2004, 11:45:24 PM »

Quote

Originally posted by beet1e
The vertical separation (can be seen from side panel in film) was 1400ft at the time I took damage. That distance is represented by side a in the triangle.

That's what I was looking for. I was wondering how you knew you had a 1400 ft separation. After that it's just simple geometry.

Thank you for the clarification.

B17Skull12 · « **Reply #18 on:** May 30, 2004, 01:10:29 AM »

Quote

Originally posted by beet1e
I made the calculations from a diagram like this.

The diagram above shows the relative positions of BillReed's Spit and my F4U. BillReed's Spit was at point A. My F4U was at point B.

The straight line distance between us is represented by side c in the triangle. 680 yards = 2040ft. The vertical separation (can be seen from side panel in film) was 1400ft at the time I took damage. That distance is represented by side a in the triangle.

We can easily calculate the value for angle B. Cosine B = a/c = 1400/2040 = 0.686. My calculator does not have a Cosine reversal function, but I was able to determine that the angle whose cosine is 0.686 is approximately 46.6°.

We know that angle C is a rightangle of 90° as side b is a horizontal line representing the horizontal distance between BillReed and myself. And we know that the three angles in a triangle add up to 180°. Therefore A = 180-B-C. Given that B= 46.6 and C= 90, A= 180-90-46.6 = 43.4°.

math i can understand.

that sounds about right. can you say lucky?

Shuckins · « **Reply #19 on:** May 31, 2004, 09:38:37 AM »

Beetle,

There's something else at work in that situation.

A hunter stalking his prey in the mountains is often faced with a similar problem.

If he has to take an uphill shot at a mountain goat, an experienced shooter would realize that the actual distance to his target is represented by line B, and will figure his bullet drop accordingly. The uphill angle makes little difference. An inexperienced shooter would overestimate the distance, hold his sights to a higher point, and shoot OVER the goats back.

Determine the distance represented by line B and you will get the actual distance to target that would effect bullet drop. He was "closer" than you think.

Regards, Shuckins/Leggern

beet1e · « **Reply #20 on:** May 31, 2004, 10:35:42 AM »

Shuckins,

I don't think I follow you. In my diagram, the upper case letters represent the angles, and the lower case letters represent the sides opposite the corresponding upper case angle. B is an angle - I assume you mean side b. Assuming that line c = 2040ft, and line a = 1400ft, we can calculate the length of line b using Pythagoras: It comes out to 1483ft. But that represents the horizontal separation between us, and NOT the actual distance between us which was 2040ft. View the film and see for yourself - 680yds.

Also remember that at a distance of 2040ft, the trajectory of the projectiles would be curved. Angle A represents the upward angle from BillReed's position to my position. I made it 43.4°. But given the curved trajectory, he would have had to pull the nose up higher still, unless we're talking laser gunnery. Which means he pulled pretty hard and the airframe would have been loaded, making the chances of an accurate shot even less likely. At least that's how it works in other games...

BenDover · « **Reply #21 on:** May 31, 2004, 10:59:07 AM »

Aren't the names of lines ment to be the angles that they are between?

E.g. Side b really should be side AC

beet1e · « **Reply #22 on:** May 31, 2004, 11:31:36 AM »

Quote

Originally posted by BenDover
Aren't the names of lines ment to be the angles that they are between?

E.g. Side b really should be side AC

Just another notation, Ben. Come to think of it, your example sounds familiar from school days. I had to revise all my trig some years ago when I wrote a flight planning application in Visual Basic. My source of reference was a Basic tutorial reference book which I bought in the US in 1979. Heavily outdated now, but it did have an excellent trig chapter, and used the notation I have used in my diagram here.

hitech · « **Reply #23 on:** May 31, 2004, 12:31:02 PM »

Beetle's angles are correct. Not sure why he didn't just use inverse sin i.e sin(A) = a/c , btw beetle your windows calculator has inverse trig functions.

But not sure what point your trying to make about 43 degs?

HiTech

Shuckins · « **Reply #24 on:** May 31, 2004, 01:03:06 PM »

Beetle,

What I'm trying to say is that the bottom line of the triangle, or b, represents the actual distance between the Spit and your Corsair, at least in terms of bullet drop. If gravity and bullet drop are modeled correctly, the shot he made on you was not as "long" as you think it was. In other words, the amount of "hold-over" for the shot would be the same as that from point A to point C along the bottom, or horizontal line, of the triangle.

Ergo, the shot was difficult, but not impossible.

Sorry for the confusion.

Shuckins/Leggern

Mugzeee · « **Reply #25 on:** May 31, 2004, 02:55:06 PM »

Quote

Originally posted by beet1e
I made the calculations from a diagram like this.

The diagram above shows the relative positions of BillReed's Spit and my F4U. BillReed's Spit was at point A. My F4U was at point B.

The straight line distance between us is represented by side c in the triangle. 680 yards = 2040ft. The vertical separation (can be seen from side panel in film) was 1400ft at the time I took damage. That distance is represented by side a in the triangle.

We can easily calculate the value for angle B. Cosine B = a/c = 1400/2040 = 0.686. My calculator does not have a Cosine reversal function, but I was able to determine that the angle whose cosine is 0.686 is approximately 46.6°.

We know that angle C is a rightangle of 90° as side b is a horizontal line representing the horizontal distance between BillReed and myself. And we know that the three angles in a triangle add up to 180°. Therefore A = 180-B-C. Given that B= 46.6 and C= 90, A= 180-90-46.6 = 43.4°.

Now factor in kenectic energy.

beet1e · « **Reply #26 on:** May 31, 2004, 03:54:25 PM »

Hitech - quite correct, I could/should have just used sine(A), not cosine(B), but... same difference. Must have been at the third glass of wine stage.

Come to think of it, when I wrote that flight planning app, almost ALL the trig was done using Sine. Had to use Cosine for crosswind effect calculation as sine(70) = sine(110), sine(80) = sine(100) etc. I didn't know about the Windows calculator inverse trig functions - will use that from now on.

I would have thought that the greater the upward angle from which the guns are fired, the more rapid would be the loss of kinetic energy. A vertical shot upwards would be in greatest defiance of gravity, and therefore lose its kinetic energy more quickly than one fired at an upward angle of 20° - am I right? So the point about the ~43° is that there would, one assumes, be considerable loss of kinetic energy, and a fair amount of drop, ie the bullets don't fly from here to the moon in a straight line. That, plus the fact that he probably had to pull hard to get the nose up, therefore loaded airframe, would have made the shot very, very difficult.

Quote

Ergo, the shot was difficult, but not impossible.

I looked at Bill's career stats - only about 200 career kills. Which kind of makes me think that luck had a greater part to play in this scenario.

straffo · « **Reply #27 on:** May 31, 2004, 05:21:25 PM »

Quote

Originally posted by beet1e
I looked at Bill's career stats - only about 200 career kills. Which kind of makes me think that luck had a greater part to play in this scenario.

It can be also the new account of a vet coming back.

Ecliptik · « **Reply #28 on:** May 31, 2004, 11:22:13 PM »

When you consider that 1400 feet is (was) about 100 feet shy of the height of the World Trade Center towers, it makes you realise how far above, vertically, Beetle really was.

The ballistics haven't changed much, if any, in AH2, but with the new tracers and smaller hit sprites, hopefully making such shots should be the one-in-a-hundred miracle hits they really were.

SpitLead · « **Reply #29 on:** June 03, 2004, 05:27:48 PM »

Beet1e,

I call that the "golden BB". It's a simple as that. Lucky shot. Sometimes ya just have to shake your head and move on. I've been pinged from beyond 1.0 before. Once I recall getting pinged at 1.5! Go figure. Between the gun conversion factor and the shell drop off not sure how that could ever happen but it does every once in a blue moon.

Spitlead.