Clocking PVC cannon speed

Fairly involved process when you don’t have a radar gun. But possible.

Procedure:

Fire projectile a total distance of 24.5ft. Record the shot with video and sound. Analyze sound waveform to find firing time and projectile impact time. Here is the setup.

To accurately model the velocity of the projectile, assume that while the projectile is in the tube, it is accelerating at a constant rate. Then assume that this velocity is maintained until it makes contact with the target (which is quite incorrect. Therefore, this models the average velocity after the projectile leaves the tube).

Total time was found by analyzing this waveform (peaks indicate the two “thuds.” Thuds are verified by listening to the waveform:

Assignment of variables:

-Any subscript of 1 will pertain to the interval when the projectile is in the tube.
-Any subscript of 2 will pertain to the interval when the projectile has left the tube.
-Any subscript of 3 will pertain to the combination of the first interval and the second interval.

Known Variables:

D1 = 10ft
D3 = 24.5ft
D2 = 14.5ft
T3 = .119 sec

Relationship between distance:

D1 + D2 = D3

Substitute velocity and time for distance:

V1T1 + V2T2 = D3

Assume that V1 = V2/2 (During constant acceleration with zero initial velocity, average velocity is exactly half of the final velocity. Since V1 represents the average velocity while in the tube and V2 represents the final velocity, V1 = V2/2).

Therefore, when substituting for V1:

T1V2/2 + V2T2 = D3

We’re trying to find V2, and we don’t know T1 or T2, so we must use some relationships to figure out what they are.

Relationship 1: T1 + T2 = T3
Relationship 2: V1T1 = D1

We must manipulate relationship 2 to a more useful form (get rid of V1):

V2T1/2 = D1
T1 = 2D1/V2

Now, going back to the original equation:

T1V2/2 + V2T2 = D3

Plug in for T1 and simplify:

2D1*V2/(2*V2) + V2T2 = D3
D1 +V2T2 = D3

Now plug in for T2:

D1 + V2(T3 – T1) = D3

Plug in for T1 again and simplify:

D1 + V2(T3 – 2D1/V2) = D3
D1 + V2T3 – 2D1 = D3

V2 = (D3 + D1)/T3

^^ That’s the equation we were looking for. Elegantly simple if you ask me. Now we can plug in actual data.

V2 = (24.5 + 10)/.119 = 290ft/sec = 198mph.

And just as a reminder, this is the average speed after leaving the tube. Therefore, the velocity when immediately leaving the tube is likely to be significantly higher than the average.

Might explain how it:

-Absolutely destroyed full soda cans
-Went inside a full bike water bottle
-Went all the way through an empty bike water bottle
-Went all the way through an unopened bottle of myoplex.

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~ by daveinthewest on August 7, 2010.

Posted in Uncategorized

3 Responses to “Clocking PVC cannon speed”

Very nice, Dave. As always. Though I do have a quick question: why do you assume the instantaneous velocity of your projectile at start will be much larger than the average velocity? I don’t see why it would slow down all that much, especially in such a short period of time. The drag on an object at high speeds (which I would say is a safe assumption for your projectile) is given by
$\displaystyle F = - \frac{1}{2} \rho v^2 C_d A$

where ${\rho}$ is the density of the fluid (air, in this case), ${v}$ is the velocity of the object, ${C_d}$ is the drag coefficient (for a smooth sphere, in this case), and ${A}$ is the cross-sectional area of the projectile. I will let ${\rho = 1.2 \frac{\text{kg}}{\text{m}^3}}$ , ${C_d = 0.1}$ , ${r = 1 \times 10^{-2} \text{m}}$ , and thus ${A = \pi r^2 = \pi \times 10^{-4} \text{m}^2}$ . Let ${k = \frac{1}{2} \rho C_d A \approx 1.88 \times 10^{-5} \frac{\text{kg}}{\text{m}}}$ . Then
$\displaystyle F = \frac{dv}{dt} = -k v^2$

We can find ${v}$ by separating variables and integrating
$\displaystyle \int \frac{dv}{v^2} = \int -kt \ dt$

$\displaystyle \frac{1}{v} = kt + C$

$\displaystyle v = \frac{1}{kt + C}$

After we’ve solved for ${C}$ by assuming ${v(0) = v_0}$ , where I let ${v_0 = 90 \frac{\text{m}}{\text{s}}}$ , we get
$\displaystyle v = \frac{1}{kt + \frac{1}{v_0}}$

Over the course of one second, we find that ${v(1) = 89.84 \frac{\text{m}}{\text{s}}}$ , and thus ${\Delta v = 0.16 \frac{\text{m}}{\text{s}}}$ , not that great of a change. Of course, this is all in theory. I imagine you have a much better idea than me about the actual physical change in the velocity from observation. And moreover, I mostly just did this in good fun and to see if LaTeX works well in Word Press. The answer to that question is: not too good! Let me try this again…

You’ll have to show me the real thing when I come home in a week. In the meantime, enjoy Chi!

Dave Darmon said this on August 7, 2010 at 7:21 pm | Reply
I just noticed that I forgot mass in the second (and subsequent) equation. If we assume a mass of 10 g, that results in a loss of 4 m/s after a 1 second period. That’s a pretty appreciable amount.

Sorry for gumming up your comments! Feel free to delete the first one. Now I know how to write LaTeX on WordPress!

Dave Darmon said this on August 7, 2010 at 7:33 pm | Reply
Haha… very nice. Well, the problem stems from your assumption about the mass of the object and potential rotational drag. The projectile only weighs about 1 or 2 grams at most with a rather large cross section.

Although, you do have a good point about the significance of the difference in speed. If I had to guess, I would say it loses about 5m/s from the time it leaves the tube to the time it hits the wall.

Also, your comment about 4m/s lost per second. That’s verrrry slow. Keep in mind the total air time was only .119 seconds, in which it only would have lost about .4 m/s.

Good commentary though. It’s always important to keep practice in check with theory.

daveinthewest said this on August 7, 2010 at 8:44 pm | Reply

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