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Artillery and Elevation
Name: Jim H.
Status: educator
Age: 40s
Location: N/A
Country: N/A
Date: Thursday, August 22, 2002
Question:
During the civil war artillery was often placed on higher points of elevation.
If a field piece like an ordinance rifle at 15 degrees elevation could
fire a projectile 4500 feet on "level ground", what would the range be if
it were on a mountain 1500 feet above a valley? Time of flight about 5 seconds.
Is there a general rule relative to increased elevation and distance gained?
Replies:
Jim -
To solve your problem you need to convert the speed of your projectile
into two
vectors - one horizontal and one vertical. (a little trig would help)
Lets make it easy... your fire your cannon ball in the horizontal from
height of
32 feet. (a falling object accelerates at 32 feet per second per second) It
would have a vertical velocity of zero when it leaves the mouth of the cannon.
At the end of two seconds its velocity would be 64 feet per second - an
average
of 32 feet per second ( 0 + 64 divided by 2). It would hit the ground in two
seconds.
If you know the horizontal velocity - let us make it 1000 feet per second
- you
can calculate the distance traveled before striking the ground after the two
seconds calculated above.... 2000 feet.
The farther you tip the cannon up from vertical, the greater the vertical
velocity at the mouth of the cannon (and the less the horizontal).
If we could keep the 1000 feet per second horizontal velocity, and tip the
cannon
up so is would have a vertical velocity of 32 feet per second... it would
take
one second before the vertical velocity would be zero. It would then
begin its
decent. If it were the same as our first example, it would fly for four
seconds... one second up, one second down to the original height of the cannon
mouth, and the two seconds we discussed earlier. In four seconds it would
travel
4000 feet horizontally.
Not a totally realistic answer, because as the vertical velocity of the
projectile increases the horizontal velocity will decrease... unless the total
velocity (the vector sum of the H and V) increases with the tilt of the
barrel.
The realism of this example is also stretched because we are ignoring air
resistance and the motion of air currents. It would work rather well on
the moon
if we used the acceleration due to gravity of about 5 feet per second per
second.
Larry Krengel
Jim,
The "rule" is that horizontal time of flight equals vertical time of flight.
The time required to go up and come down, ending up 1500 feet below launch,
equals the time spent moving horizontally. Vertical motion has constant
downward acceleration: y=y0+{v0y)t-(1/2)gt^2. Discover the time in flight
based on initial vertical velocity. Horizontal motion has constant
velocity: x=x0+(v0x)t. V0x and v0y are the x and y components of the
initial velocity. Analyze the level ground situation to discover what
initial velocity is. It will be the same when launched from increased
elevation.
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
If the shell leaves the gun moving at an angle of 15 degrees, it will arrive
back at the original height also moving at an angle of 15 degrees. If
the
gun's height is not too much greater than the target's height, the rest
of the
shell's trajectory will be a small section of a parabola which you can
approximate as a line inclined at 15 degrees.
Tim Mooney
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