Measuring the speed of a projectile

Other Hardware Science and Measurement
1

My school has a vacuum cannon and it fires small table tennis balls. We can only calculate the velocity theoretically because my school does not have any device that is capable of measuring a projectile traveling at around 350 to 400 m/s

I was thinking of using two tin foils and measuring the time interval of opening circuit. (Using a photo resistor would be too slow)
I did some research and I did find some useful resource: http://forum.arduino.cc/index.php/topic,114435.0.html

Please give me some advice on how I should tackle this problem. (Should I just use the method I found?)

2

Photoresistor would be fine - you have the same delay in detecting at two points, so they cancel out.
interval time = (T2+time_delay) - (T1+time_delay) = T2 +time_delay - T1 - time_delay = T2 - T1.
Also has the benefit of no tin foil impacting the time by slowing down the ball at all.

3

We can only calculate the velocity theoretically because my school does not have any device that is capable of measuring a projectile traveling at around 350 to 400 m/s

That fast tennis balls - a joke :slight_smile:

4

of measuring a projectile traveling at around 350 to 400 m/s

Mach 1+ ping-pong balls?

Some serious over-estimation there, I think.

5

you can measure the soundpulse of the cannon firing and the soundpulse when it drops the floor or it hits the wall.
That too will give you two timestamps and a distance.

If you shoot right up in the air you can even determine the velocity of the ball leaving the canon as the ball makes (almost) a perfect parabola.
Gravity will decrease the velocity until it reaches zero

V = V0 - g x t

V = 0 when it is at the top, => V0 = g x t

g = 9.80665 (depends on where you live), and t is half the time between firing and the moment it hits earth again.

example:
so if t = 14.35 seconds => 0.5 t = 7.17 => V0 = g x t = 9.81 x 7.17 = 70.34 m/s

The height the ball would come is 0.5 x g x t2 = 0.5 x 9.81 x 7.172 = 252.16 meter (so don't test this indoors :wink:

The above calculations do not include drag / friction.

6

AWOL:

of measuring a projectile traveling at around 350 to 400 m/s

Mach 1+ ping-pong balls?

Some serious over-estimation there, I think.

It is a 2 meters long vacuum cannon.
P = Atmospherics pressure = 101325 Pascal
r = 0.02 meters
A = (pi)(r)^2

F = PA
F = (101325)(3.14 X 0.04) = 12732.8750
W = Fd = (12732.8750)(2) = 25465.75
W = (1/2)(m)(V)^2

V = square root((2 X 25465.75)/0.04) = 1128
V = 1128 m/s

Considering energy loss, we are just assuming it to be around 350 m/s to 400 m/s
so around 780 mph to 895 mph

7

Don't be ridiculous.

8

CrossRoads:
Photoresistor would be fine - you have the same delay in detecting at two points, so they cancel out.
interval time = (T2+time_delay) - (T1+time_delay) = T2 +time_delay - T1 - time_delay = T2 - T1.
Also has the benefit of no tin foil impacting the time by slowing down the ball at all.

I am just worried because I have never used a photoresistor for anything... and the projectile is moving very fast...
Also the photoresistor I have is a crap one (I believe) that just came with SparkFun Arduino starter kit...

Edit:
Based on simple calculation
Assuming it is traveling it at 350 m/s, a large projectile (cylinder shape with hemispheric nose and tail) which is around 6 cm
t = d/v = 0.06/350 = 0.0001714 seconds...
Arduino has 16 million clock cycle which is plenty to get readings but I think the photoresistor is just not that responsive...

9

Use a photodiode or phototransistor.

(your calculations assume a perfect vacuum on one side of the projectile)

Edit:
Sanity check.
0 - 340ms-1 implies a steady acceleration of v2 /2s = 3402 / 4 = 28 900ms-2 or about 2950g.
Can a ping-pong ball really sustain that?

10

www.phys.csuchico.edu/~lbuchholtz/Papers/VC_AJP.pdf

The above paper says it could be max 287m/s and 4700g (measured with ping-pong balls and photo gates 125-250m/s typically for 0.5 -2m cannon lengths).

11

AWOL:
Use a photodiode or phototransistor.

Do you think this Miniature Solar Cell - BPW34 will do the trick?

Datasheet:

12

Do you think this Miniature Solar Cell - BPW34 will do the trick?

No, I don't think it'll be fast enough

13

AWOL:

Do you think this Miniature Solar Cell - BPW34 will do the trick?

No, I don't think it'll be fast enough

Oh... I don't know what device to use now...

14

I do believe I mentioned a photodiode or a phototransistor.
If you google "ballistic computer", you may even find circuits.

15

I do believe I mentioned a photodiode or a phototransistor.

That's what you need, a photodiode or a phototransistor. I don't know why you want to use something different. They aren't that difficult to find.

Can a ping-pong ball really sustain that?

I have the same impression. If the speed was to be that much, the ping pong ball will probably disintegrate inside the cannon.

16

It is a 2 meters long vacuum cannon.
P = Atmospherics pressure = 101325 Pascal
r = 0.02 meters
A = (pi)(r)^2

F = PA
F = (101325)(3.14 X 0.04) = 12732.8750
W = Fd = (12732.8750)(2) = 25465.75
W = (1/2)(m)(V)^2

V = square root((2 X 25465.75)/0.04) = 1128
V = 1128 m/s

Redo your calcs, this time square r, F=(101325)(3.14x0.02x0.02)=127.328
W=127.328x2=254.657
V=sqrt((2x254.657)/0.04)=112.83 m/s I think

Tom

17

Well spotted!

18

I hope I got it right, it was 10 to 1am here when I did the calc.
Tom

19

I don't think it is possible to fire a ping pong ball at 300 metres per second.

20

F = PA
F = (101325)(3.14 X 0.04) = 12732.8750

So the 3.14 x 0.04 which you have there, is supposed to be the cross-sectional area of the bore of the cannon,
calculated by pi * r^2 ?

and r is 0.02 m = 2 cm ?

and r squared is ???