Help with science project to launch a projectile

[SSKscioly]

TL;DR Summary: Hi, I am a high schooler competing in a science competition building a trajectory to shoot the projectile similar to the one in the thread below.

https://www.physicsforums.com/threa...launch-projectile-science-competition.773745/

The 4 inch drop tower is connected to the 1.5 inch pvc that shoots the ping pong ball projectile. I am dropping a mass of 2.5 kg weight in to the drop tower that acts as the piston. The goal of the competition is for the projectile to land anywhere between 2 and 8 meters. No matter what the mass of the piston or air tight it is in the drop tower , my projectile is not launching past 5m. i am able to launch the ball to the same distance using a 3d printed weight (around 120 gms). So basically having heavier mass is not doing anything. I am at a loss as no matter what i tried or changed i can't seem to go past 5m. Please help on what i am doing wrong,

 

[kuruman]

Please help on what i am doing wrong,

You say that "The goal of the competition is for the projectile to land anywhere between 2 and 8 meters." Your projectile lands at 5 m. What makes you think that you are doing something wrong if the projectile apparently meets the goal of the competition? You might wish to see what happens if you vary the angle of projection relative to the horizontal.

If you require additional help, it would be a good idea to post a photo of your contraption.

 

[SSKscioly]

You say that "The goal of the competition is for the projectile to land anywhere between 2 and 8 meters." Your projectile lands at 5 m. What makes you think that you are doing something wrong if the projectile apparently meets the goal of the competition? You might wish to see what happens if you vary the angle of projection relative to the horizontal.

Hi Kuruman, Thank you so much for responding Sorry for not clearly mentioning. The near and far targets can be placed anywhere between 2 and 8 meters. That means the far target can be at 8m. We would not know the targets until right before the competition. So, the goal is for the projectile to shoot past 8 meters so I can then play with various heights and angles to land on different distance targets. Right now my 4 inch drop tower is 28 inches tall from the floor and the barrel that shoots the projectile at 24 inches long.

I am confused on why the mass of the weight being dropped not impacting the air pressure in the 4 inch chamber? I get the same distance using lighter 3d printed mass (120 gms) vs 2.5 kg lead puck. Shouldn't heavier weight have more gravitational pressure which should give more force to the ping pong ball to shoot further?

 

[kuruman]

How freely does the ping-pong ball move through the launching tube from beginning to end? At first it must provide some resistance through friction to let the pressure build up behind it. Once it gets going however, it must move with very little resistance from the tube.

 

[hutchphd]

Where, in the drop tube, is the heavy mass when the ball exits the barrel of the cannon tube? That will tell you a lot about the dynamics (put a thin stick on it as an indicator and compare the heavy and light weights) Video would be good (can you share?)

 

[SSKscioly]

Thanks Kuruman, the ping pong ball diameter is 40 mm and when the ball is dropped in the barrel it is pretty snug at the elbow. It does take a little force if i need to wiggle it out but once the mass is dropped, it shoots freely. That's why I'm confused on why it isn't able to shoot that far since it fits perfectly in the barrel and shoots out freely. Here's the video that inspired my design:

 

[SSKscioly]

Where, in the drop tube, is the heavy mass when the ball exits the barrel of the cannon tube? That will tell you a lot about the dynamics (put a thin stick on it as an indicator and compare the heavy and light weights) Video would be good (can you share?)

Hi Hutchphd, please excuse my illustration. I placed a wood block at the bottom where the mass falls, so it wouldn't hit the floor (highlighted in blue in the attached picture). The weight(highlighted in red) falls exactly near the elbow and covers the hole of the elbow by 3/4 once it is at the bottom. Both the light and heavy mass are identical in height and falls to the exact position

 

[Lnewqban]

No matter what the mass of the piston or air tight it is in the drop tower , my projectile is not launching past 5m. i am able to launch the ball to the same distance using a 3d printed weight (around 120 gms). So basically having heavier mass is not doing anything.

My suggestions to reduce air leaks as much as possible:

Seal well between the base of the column and the floor.
Use Teflon tape to seal all the threaded couplings, even the one that pivots to adjust the launching angle.
Make several grooves around the perimeter of the piston, in order to create a labyrinthic seal that reduces the volume of air that escapes while the piston drops. make a chamfer at the edges. Make the piston as tall as possible.

Please, see:
http://docs.codecalculation.com/mechanical/seals/labyrinth-leakage.html

 

[hutchphd]

I think the issue may be more fundamental. Suppose the backpressure caused by the pingpong ball is negligible. Then the time course of the falling piston will be the same for any driving weight. This will mean that the time course of the projectile will also not depend upon the piston mass.
So do the different mass pistons fall the same, or does the heavy one fall faster?
Does it matter whether there is a ball in the barrel?
These answers will be indicative of the problem

 

[SSKscioly]

My suggestions to reduce air leaks as much as possible:

Seal well between the base of the column and the floor.
Use Teflon tape to seal all the threaded couplings, even the one that pivots to adjust the launching angle.
Make several grooves around the perimeter of the piston, in order to create a labyrinthic seal that reduces the volume of air that escapes while the piston drops. make a chamfer at the edges. Make the piston as tall as possible.

Please, see:
http://docs.codecalculation.com/mechanical/seals/labyrinth-leakage.html

Hi Lnewqban, Thanks you so much for responding. I sealed the joints of the barrel and the drop tower with the pvc cement so there are no air leaks. i guess this is where I am getting confused on what it can be. How can I get the labyrinth seal on a lead puck?

 

[erobz]

I think we can ballpark the dynamics by solving the following system of equations: Since the flows are well below Mach 0.3 we can use an incompressible flow approximation and apply Bernoulli's:

$P_1$ is the pressure acting on the falling piston
$v_1$ is the velocity of the falling piston
$P_2$ is the pressure acting on the projectile
$v_2$ is the velocity of the projectile.

Applying Bernoullis's:

$$ \frac{P_1 - P_2}{\rho} = \frac{v_2^2 - v_1^2}{2}$$

Then applying Newtons law to the dropping mass $M$ ( $\downarrow^+$):

$$-P_1 A_1 + Mg = M \dot v_1 $$

And simultaneously to the projectile of mass $m$ ( $\nearrow^+$):

$$ P_2 A_2 - mg\sin \theta = m \dot v_2 $$

And lastly continuity relates $\dot v_1$, and $\dot v_2$ by

$$ v_1 A_1 = v_2 A_2 \implies \dot v_1 A_1 = \dot v_2 A_2$$

I think that leads you to a (solvable) non-linear ODE in $v_2$ (the velocity of the projectile).

You should be able to use that to investigate the performance characteristics of different weights $M$. Thats not going to fix your problem, but it could shed light on the basic performance behavior.

 

[Lnewqban]

Hi Lnewqban, Thanks you so much for responding. I sealed the joints of the barrel and the drop tower with the pvc cement so there are no air leaks. i guess this is where I am getting confused on what it can be. How can I get the labyrinth seal on a lead puck?

What about the seal between the base of the column and the floor?
How did you make the puck?
Do the rules give you liberty to change the lengths of both pipes?

 

[SSKscioly]

What about the seal between the base of the column and the floor?
How did you make the puck?
Do the rules give you liberty to change the lengths of both pipes?

The only rule in regards to the height is the whole device has to fit in 75 x 75 cm cube. I purchased the lead puck that weighs 5 lbs and 3.5 inch in diameter. I used the tape to make it as air tight as possible to match the diameter of the 4 inch drop tower. I put the 4 inch drop tower pipe on a 4 inch tee and put a wooden block (highligted in green) on the top of the tee just under the elbow to seal the pipe and the floor. So basically everything is air tight

 

[Lnewqban]

The only rule in regards to the height is the whole device has to fit in 75 x 75 cm cube. I purchased the lead puck that weighs 5 lbs and 3.5 inch in diameter.

Then, as the internal diameter of a schedule 40 4-inch PVC pipe is 101.5 mm, you had to build up the radius of the puck with 6 mm of tape around its perimeter.
The height of the puck should be around 1-1/4 inch.
I would use two or three narrower strips of tape to create the seal.

Perhaps the puck is not perfectly cylindrical and air leaks between it and the 4-inch pipe.
Could you test the time descending while the small pipe is capped and compare it to the time of normal operation?

 

[erobz]

If solving the resulting ODE is not an option you want to try, then you can still examine a theoretical maximum (agreeing with the solution of the ODE at least) projectile velocity with basic kinematic relationships by assuming incompressible flow:

$$A_2 v_2 = A_1 v_1 $$

You might do this to verify if there is at least chance of hitting your target distance with the design.

If there were absolutely no pressure or drag/friction forces acting on the falling piston, can you write the velocity $v_1$( piston), and $v_2$(projectile) as a function of time $t$(working under the assumption above of incompressible flow)?

 

[hutchphd]

I think I see a (probably the ) problem. You have not reproduced the initial gravity fed (award winning) choices. As @erobz points out consider the case of incompressible perfect massless air. The condition of incompressibiliy $$A_1 v_1=A_2 v_2$$ and so the muzzle velocity for the very light ($m\ll M$) ping pong ball will be constrained by the speed of the puck. $$ v_1^2\lt 2 g h_1$$ $$h_1=\frac {A_2} {A_1} L $$where L is the Length of barrel (2). And so $$ v_2 =\frac {A_1} {A_2} v_1$$ $$ v_2 \lt \sqrt {\frac {A_1} {A_2} 2gL} $$
Using this analysis should allow calculation of the ~max range
Strangely (perhaps) this gun may work better if it is a little leaky, allowing the puck to gain speed before impulsively interacting with the projectile. You need some upper holes in the vertical tube (or a bladder driving the gun barrel). A real analysis using the compressability of air would be interesting.

 

[erobz]

I think I see a (probably the ) problem. You have not reproduced the initial gravity fed (award winning) choices. As @erobz points out consider the case of incompressible perfect massless air. The condition of incompressibiliy $$A_1 v_1=A_2 v_2$$ and so the muzzle velocity for the very light ($m\ll M$) ping pong ball will be constrained by the speed of the puck. $$ v_1^2\lt 2 g h_1$$ $$h_1=\frac {A_2} {A_1} L $$where L is the Length of barrel (2). And so $$ v_2 =\frac {A_1} {A_2} v_1$$ $$ v_2 \lt \sqrt {\frac {A_1} {A_2} 2gL} $$
Using this analysis should allow calculation of the ~max range
Strangely (perhaps) this gun may work better if it is a little leaky, allowing the puck to gain speed before impulsively interacting with the projectile. You need some upper holes in the vertical tube (or a bladder driving the gun barrel). A real analysis using the compressability of air would be interesting.

Yeah, it's funny. The holes allow the puck to accelerate to a higher initial velocity before it really begins to accelerate the projectile. Thus, reaching the lower section (where there are no holes) at a higher speed to transfer the momentum/energy to the projectile in a short intense blow. Neat!

 

[erobz]

So, since this is beyond the scope of the high school analysis, (without the holes, and neglecting compressibility effects) If you solve the system of equations in #11 you get the following ODE w.r.t. the coordinate of the projectile $x_2$:

$$ - \left( \frac{M r A_2+m A_1}{A_1A_2} \right) \frac{dv_2}{d x_2} v_2 = \frac{\rho}{2}\left(1-r^2 \right)v_2^2 + \lambda \tag{1} $$

Where
$M$ is the mass of the piston
$m$ is the mass of the projectile
$r$ is the ratio of areas $\frac{A_2}{A_1}$
$\rho$ is the density of air
and

$\lambda = \frac{mg \sin \theta}{A_2}- \frac{Mg}{A_1}$ ( a constant parameterized by $M$)

$\theta$ is the angle the barrel makes with horizontal.

The solution to (1) is given by:

$$ v_2 = \sqrt{ \frac{2}{\rho} \frac{1}{1-r^2} \lambda \left( e^{ \frac{-A_1 A_2 \rho ( 1-r^2) }{M r A_2+m A_1} L} - 1 \right) } $$

Where $L$ is the length of the barrel.

Here is a plot of $v_2$ vs. $M$ for what I thought are reasonable parameters:

That horizontal asymptote is found via the relationship @hutchphd showed.

 

[SSKscioly]

Thank you @hutchphd and @erobz for the guidance. I will try drilling holes and will test tonight.

 

[erobz]

Thank you @hutchphd and @erobz for the guidance. I will try drilling holes and will test tonight.

I would drill, test,drill…it should need some sealed volume at the lower end.

 

[SSKscioly]

I drilled the holes and now the ball shoots around 5.5 meters. Do I need to drill more holes?

 

[erobz]

I drilled the holes and now the ball shoots around 5.5 meters. Do I need to drill more holes?

How far down the pipe and how many holes did you drill? A picture of updates would be good.

 

[SSKscioly]

Sorry, I tried uploading the video and it is too big.. I drilled 30 holes all around the tube. and above 12 inches from where the weight drops to the bottom

 

[erobz]

Sorry, I tried uploading the video and it is too big..

just a picture would be fine.

I drilled 30 holes all around the tube. and above 12 inches from where the weight drops to the bottom

Well, 30 sounds like a good bit. What diameter? Just make sure you save enough sealed volume above the barrel port to fully displace a barrel volume with the piston. Can you calculate how high the holes should start above the barrel port? Below that would probably be detrimental to performance. I would still go toward there in stages, even after the computation.

 

[erobz]

Also, make sure you have de-burred the holes if you haven’t done so already.

 

[hutchphd]

I think I see a (probably the ) problem. You have not reproduced the initial gravity fed (award winning) choices. As @erobz points out consider the case of incompressible perfect massless air. Let the dropped weight have mass M and coordinate z from base and the projectile mass m and coordinate x down the barrel from the base. The condition of incompressibiliy $$A_1 v_1=A_2 v_2$$ and so the muzzle velocity for the very light ##m(ping pong ball will be constrained by the speed of the puck. $$ v_1^2\lt 2 g h_1$$ $$h_1=\frac {A_2} {A_1} L $$where L is the Length of barrel (2). And so $$ v_2 =\frac {A_1} {A_2} v_1$$ $$ v_2 \lt \sqrt \frac {A_1} {A_2} 2gL $$

Well, 30 sounds like a good bit. What diameter? Just make sure you save enough sealed volume above the barrel port to fully evacuate a barrel volume. Can you calculate that?

I a;so have a practical tip. When prototyping with PVC I glue it together lightly (if glue is really requiired ) with pure silicon caulk. Easy to disassemble and modify as necessary.
As mentioned the sealed volume at the upright tube needs to be slightly bigger than the active volume of the barrel (breech to tip). Also I think adjusting the angle of the barrel will give you finer control than the "size of the charge" (i.e weight and altitude of puck)
Send your calibration data! Good Luck.

 

[SSKscioly]

Hi @erobz,@hutchphd , I am totally lost. No matter what i did, sanded smooth inside, added lubrication, changed the angle of the barrel, I am not able to shoot past 6 meters now. This is how it looks now before attaching the barrel. You think this design may not work for what I am trying to achieve?

 

[erobz]

Hi @erobz,@hutchphd , I am totally lost. No matter what i did, sanded smooth inside, added lubrication, changed the angle of the barrel, I am not able to shoot past 6 meters now. This is how it looks now before attaching the barrel. You think this design may not work for what I am trying to achieve?

View attachment 335345

But you have in fact improved the range, correct? Have you calculated the internal working volume of the barrel to determine how much of the tower must not have holes in it above the top of the barrel port?

 

[SSKscioly]

Hi @ erobz ,

The total volume of the barrel is 301.59 cubic inches ( 24 inches height and 4 inch diameter)

The sealed chamber is 9 inches down from where the holes to the top of the weight when it is dropped , so if I am calculating correctly, my internal working volume is 113.10 cubic inches

 

[hutchphd]

You need data to determine what is going on. I suggest firing the ball straight up for tests (one less variable to worry about). Measure (video against a marked board?) the max height of of the ball for the two different weights at five different drop altitudes. Record the results and send them along with analysis (maybe a graph)?. We can probably help.
The difference between "doing science" and "playing" is that for science you quantify and record all results!......
Also how high does the ball need to go in these tests to get the range you need when you tilt the barrel? You should be able to calculate this (ignoring air resistance)

 

[erobz]

Hi @ erobz ,

The total volume of the barrel is 301.59 cubic inches ( 24 inches height and 4 inch diameter)

The sealed chamber is 9 inches down from where the holes to the top of the weight when it is dropped , so if I am calculating correctly, my internal working volume is 113.10 cubic inches

Your barrel is 1.5 inch diameter. What is its length?

 

[SSKscioly]

24 inches

 

[erobz]

24 inches

Then what is the internal volume of the barrel?

 

[SSKscioly]

42.41 cubic inches

 

[erobz]

You are trying to figure out where the red dashed line is. That is the line you should not drill holes below. You are doing that by comparing the volume inside the barrel, to the volume inside the tower between those lines. we need enough air inside the sealed part of the tower to completely displace the air in the barrel.