nitroboy
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Parabolic Nature of Projectile Motion.
I am currently investigating the nature of projectile motion. I'm not using a cannon or anything such as that, but, unlike the usual, something
safe that can represent the motion.
I am filling tubes of PVC with water and drilling holes in the sides and investigating the length of the arc in relation to the water pressure behind
it, the size of the hole, and the height of the hole.
So if anybody thinks they might be able to help me out with any of this, it would be much appreciated.
In the mean time back to learning the maths i need for this thing.
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darkflame89
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Is this chemistry? Anyway, i know projectiles are affected by their weight, angle it was shot from, force given to it. I do not think that the size of
the hole has anything to do with the projectile strength.
Ignis ubique latet, naturam amplectitur omnem.
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Proteios
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to a first degree approx. it should be parabolic. However, the more detail you look at the more complicated the problem.
The problems you have are viscosity from the air, of which atm. pressure and humidity are factors, and interal drag from non-linear flow of water
going through a hole.
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Chemtastic
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I solved a problem like this a couple years ago...it was a lot harder than it looked...
A cannon which fires a projectile at velocity v is located x meters from a wall with a target y
meters up that [vertical] wall. At what two angles can the cannon be fired in order to hit the target?
Like I said, this was quite a bit harder than it looked...I filled 3 sides of paper...then again, I have a penchant for taking the long way .
If anyone wants to try to solve this problem, two things. First, if the cannon's velocity is too low, there's no solution, and if it's
just perfect, there's exactly one solution. Second, I took the parametric-then-trig identities-then-combined sin-cos summation route; if anyone
actually solves something easier, I'd like to see it.
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Magpie
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solve for angle theta
Eqn 1: X=VCos(theta)t
Eqn 2: Y=VSin(theta)t-(1/2)gt^2
where t = time of travel, and g = acceleration due to gravity.
Rearranging Eqn 1, t=X/[VCos(theta)].
Substitute this in Eqn 2 for t, i.e.,
y=xtan(theta) -(1/2)g[x/(VCos(theta))]^2
Solve for theta by trial and error unless there is a more direct route. Graphing would likely speed the T&E solution.
There will likely be 2 solutions. This would correspond to hitting the target on the rise, or on the fall, of the trajectory I would guess. Also V
would have to be adequate to get it to the target, as previously stated.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Geomancer
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Call the hieght of the place the cannon is pointed at on the wall t=x tan a where a is the desired angle. Since the amount the projectile falls from a
straight line is linear with respect to the square of the flight time, and the flight time with gravity is equal to that without, the point of impact
is given by t-c(x^2+t^2), with c being some physics stuff, probably g/(2v^2). Use the quadratic equation to find solutions in terms of t, and convert
back to angles at the end.
Nitroboy: I like the experiment. Try and make the hole nice and smooth to minimize viscous energy loss.
[Edited on 4-7-2004 by Geomancer]
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unionised
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Am I missing something here?
"I'm not using a cannon or anything such as that"
"A cannon which fires a projectile at velocity v is located x meters from a wall "
"Call the hieght of the place the cannon is pointed at on the wall "
If you drill holes in the side of a (thin walled) can full of water the water comes out horizontally. It is not common practice to fire cannon
horizontally.
Once it leaves the can the water will follow a nearly parabolic path, air resistance will have some effect on this but it's still a good
aproximation.
The velocity at which the water leaves the hole is dependant on the viscosity but, if the hole is big, this effect is small and then the velocity can
be calculated by the application of Bernoulli's equation or the conservation of energy (they amount to the same thing, and I'm sure I got
the spelling right for conservation of energy).
In theory you probably have to account for surface tension too but I think this will be too small to matter.
You might want to look at the effect of viscosity by using water at different temperatures, this affects the viscosity rather more than the density.
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Magpie
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Torricelli's Law
Getting back to nitroboy's question:
Torricelli's Law, which likely reduces from Bernoulli's equation, says
V=(2gh)^(1/2) where V=the exit velocity of the liquid, g the acceleration due to gravity, and h is the height of liquid above the hole.
So with this equation you solve for the velocity directly without even having to know what liquid you are using! Of course this neglects the effects
of viscosity which may or may not be significant - with water likely not. Since you are saying you have a horizontal hole then theta = 0. You have
computed V directly. The horizontal distance, x, should then be computed from my equation in the post above. y is the distance from the hole to the
bottom of the pipe (a negative number).
If I did my algebra right, using only the positive root, this reduces to:
x =2[(-y/h)^(1/2)]
It is interesting that this final result does not incorporate either the density of the liquid or the acceleration of gravity.
The single most important condition for a successful synthesis is good mixing - Nicodem
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nitroboy
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Sorry it's been so long, but here is my results.
Test 1 (Variable Height) (aperture = 4mm)
Height(cm) - Distance of Parabola(cm)
90 - 45.5
80 - 64.5
70 - 84
60 - 89
50 - 75
40 - 100
30 - 100
20 - 88.5
10 - 62
est 2 (variable Aperture width)
Width of hole (mm) - Distance (cm)
4.5 - 88.5
5 - 90
5.5 - 91
6 - 92.5
6.5 - (untested)
7 - 53
7.5 - 94.5
8 - 94.5
8.5 - 92.5
9 -89
9.5 -94.5
10 - 93
The problem i am having is with understanding the relationship with height (test 1) and pressure (test 2). I see that exit velocity is eqaul to
(hg)^(1/2), but it does not take into account pressure, which i don't understand at all.
[Edited on 3-8-2004 by nitroboy]
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Magpie
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Look at the Bernoulli equation found in most physics and fluid mechanics texts. It expresses fluid energy in head of fluid, or the unit of length.
In the Bernoulli equation height (head), pressure, and velocity of the fluid, respectively, can all be expressed in units of head (length).
For pressure convert the height to equivalent pressure by multiplying by the fluid density, e.g.,
ft = ft(pounds force/ft3) = lbf/ft2
lbf/ft2(1 ft2/144 in2) = psi
So for water at 34 ft height =
34 ft(62.4 lbf/ft3)(1 ft2/144 in2) = 14.7 psi = 1 atm
(My apologies to non-US members for using these archaic units.)
[Edited on 4-8-2004 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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JC
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The viscosity isn't important, if you are using the velocity of the liquid, since the losses are already taken out. Air resistance is negligble,
since the air has already been moved out the way by the previous water, assuming you have let it get to a fairly steady state.
I always start by resolving the velocity into the x and y (and z if any) so you can work things out using the basic equations of motion.
The time of flight times the horizontal component of the velocity will tell you how far it will go. You can easily work backwards, since you are
starting at horizontal (unless you steer a nozzle), and you can measure how far it goes. After all, measuring the velocity of the water directly is
really hard!
You can also work out the time of flight by working out the time taken for the water to hit the floor, since the only force acting on it in the
vertical is gravity, with the usual acceleration. You might find it best to work from this to determine the velocity, via the distance travelled in
that time.
Hope this helps.
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