From: Joseph Harper
Sent: Wednesday, January 15, 1997 7:52 PM
To: Joseph K. Huffman (Volt Comp)
Subject: RE: Six at once.

After looking over the calculations, I would have to assume that you don't have a television to vegetate in front of while you are staying here during the week, eh? <grin>

An additional item that throws a kink in your calcuations is the fact that several of your first pieces of brass bounced off both my hand and the timer which deflected them back into the air. At that point, they had to again lose velocity before accelerating under the influence of gravity. This would keep the earlier pieces of brass still in play while the rest of the brass was ejected into this system equation. These later pieces of brass would have the standard trajectory you describe in your formulas below since by that time I was able to move my hand and avoid serving as a vector redirector.

And folks wonder why I hat with a broad rim. <grin>

-- RevolverMan
Team Flame

----------
From: Joseph K. Huffman (Volt Comp)
Sent: Wednesday, January 15, 1997 4:35 PM
To: Joseph Harper
Subject: Six at once.

I wondered about your statement about there being six rounds of brass in the air at once. Could it be true? Particularly if I had good hits on the target (my 2.30 S string). So, I did some number crunching. I am a bit skeptical, but it is possible -- if the brass is ejected to a height of 9.5 feet or more. Or, if a more reasonable figure of 7 feet is used and the time from the first shot until the last is less than or equal to 1.009 seconds (my faster, less accurate strings)....

Of course at an indoor range in the booth we have brass bouncing off the walls and ceiling "ornaments" so it is mostly a moot question, but it does have application outdoors...

Boring details follow:

-joe-
--------------------------
The equation of motion for a body under the influence of gravity with an initial velocity, assuming negligible wind resistance and ignoring all but the vertical component is (I would derive it, but just take my word for it):

y == y0 + v0 * t - 1/2 * 32.2 * t^2 Where:
y is the current position above the ground in feet
y0 is the initial position
v0 is the initial velocity in feet per second.
t is time in seconds.

Assume:

y0 is the ejection port of the Ruger P89 in my hands at about 5 feet above the ground.

v0 is unknown, but we can figure it out from the maximum height that the brass attains, say 7 feet above the ground. Which we will do now:

Assume y = 7 feet, find v0, such that this is the maxim y.

The equation of velocity (using the same assumptions) is:

v == v0 - 32.2 * t

At the maximum height, v = 0.

So

0 == v0 - 32.2 * t

or

t == v0/32.2

Substituting for y == 7 we get:

7 == 5 + v0 * t - 16.1 * t^2

Substituting for t == v0/32.2

7 == 5 + v0 * (v0/32.2) - 16.1 * (vo/32.2)^2

Solving for v0:

2 == v0^2 (1/32.2 - 16.1/(32.2)^2)
v0^2 == 2/(1/32.2 - 16.1/(32.2)^2)
v0^2 == 128.8
v0 == 11.35 fps

Doing all the proper substituting in the first equation:

y == 5.0 + 11.35 * t - 1/2 * 32.2 * t^2

Solving for t when y == 0 (ground level):

0 == 5.0 + 11.35 * t - 1/2 * 32.2 * t^2
t = (-11.25 +/-(11.35^2 - 4 * (-16.1 * 5))^(1/2))/(2 * -16.1)
t = 0.3494 +/- 0.6594

Since the are only concerned with time in the forward direction, the answer is:

t = 1.009 seconds elapse from the time of ejection until the time of impact on the ground. Assuming my first shot took 1.0 seconds and the remaining five shots were evenly distributed, my 2.30 string would be composed of shots at 1.0, 1.26, 1.52, 1.78, 2.04, 2.30. Subtracting 1.0 from each of the shots (the time the first round of brass became airborne) we get, 0.0, 0.26, 0.52, 0.78, 1.04, and 1.3.

So, if my assumption of a max altitude of 7 feet is correct, and the first shot was at 1.0 seconds, there were at most 5 rounds of brass in the air at once. Working backwards to find the max altitude to satisfy the 6 rounds in the air at once we get:

0 == 5.0 + v0 * 1.3 - 16.1 * (1.3)^2
v0 == 17.08 fps

v == v0 - 32.2 * t
0 == 17.08 - 32.2 * t (at ymax)
t == 0.5306

ymax == 5.0 + 17.08 * 0.5306 - 16.1 * (0.5306)^2
ymax == 9.53 feet.