# Chronograph (bullet velocity) - Help plz

Discussion in 'The Projects Forum' started by Jacob J, Aug 15, 2009.

1. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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Hello

I need some help. I have come up with an idea of making my own chronograph, so I can measure the bullet velocity, when I shoot a bullet out of my hunting rifle.

I am not very good at any electronic stuff, though I have made some simple circuits, like a PWM circuit (also with help some you guys).

I have searched the net and found some informations about the construction of such a device and here is what Ive found.

I make two frames 4x4 inchs, where there are four IR leds in the buttom and 4 photodiodes in the bottom. I flood both frames with IR light and the photodiodes pick up the light. When the bullet is passing through the frames it works like a switch. Although Ive learned that I must use a NAND timer to send the signal from the frames into and that then works as a switch. I havnt been able the figuere out how I must proceed.

I would like to be able to read the result on something, I have been thinking about segment blocks, but if there is an easier way, I would like to do that.

Can you please tell me how I must go on with this project?

And please, it must be as simple as it can be. But I also just want to be able to read the velocity or some result I then can do the math on and get a result.

Hope you all can help me

Best regards

Jacob Jensen (Denmark)

2. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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Ive found out, that the devices you buy uses a crystal clock/osciliator to measure the time, but how do I put it all together?

My idea; Each frame with the IR leds and photodiodes sends a signal into an TTL NAND input. Then the TTL switches on a crystal clock on the first signal from frame 1 and the TTL switches the crystal clock off again, when it gets a signal from frame 2.

Thats the way I think it should work, but how do I put it together and how can I get a reading of the results?

Apr 20, 2004
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4. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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That looks like a very very complex setup. I was looking for something much more simple, but thanks for the link, I havnt seen that project before.

5. ### beenthere Retired Moderator

Apr 20, 2004
15,815
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The device needs a means of turning on and then off. It needs a clock source and counter to record the duration between the on and off event. And it need to be able to display that time, plus a manual clearing so the caount can be made 0 before the next test.

The complexity of that circuit is only superficial. That is pretty minimal as to circuitry. You might use fewer IC's by using a microcontroller, but I doubt it it would save many.

6. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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Hmm ive found this very very simple construction. Would it work?

"The simplest version I can imagine would be the following:

a.) a battery
b.) a resistor and capacitor in series across the battery, with a time constant (R*C seconds, where R is in ohms and C in Farads) chosen to be about 10 times longer than the time you expect the bullet to need to travel between the sensors
c.) two fine wire sensors that are broken by the passage of the bullet
d.) a voltmeter reading the voltage across the capacitor (should be very high impedance, so as not to drain the capacitor faster than you can read the voltage).

The first fine wire is across the capacitor, shorting it to zero volts while it is in place, but allowing the capacitor to charge once it's broken. The second fine wire is in series with the capacitor (for example, between capacitor and resistor); when the second wire breaks, the capacitor stops charging. You read the voltage quickly after the bullet passes through both wires.

Since you are timing an interval small compared to the time constant, you can treat the current as roughly constant with I = V/R, where V is the battery voltage. The voltage Vc across the capacitor will be Vc = (I*t)/C, where t is the time between the sensors. Combining the first two equations, you get Vc = V ( t / R*C ). It is important to have t << RC for this formula to be accurate. You can turn the formula around to caculate the time by t = (R*C) * (Vc / V ) . And finally, the velocity is v = d / t, where d is the distance between the sensors.

There is also a more precise calculation using the proper exponential equation for the voltage across the capacitor if you want to get fancy."

7. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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How do you find that time-constant? Lets say I have a 8,2k resistor laying around, what capasitor would I need? I assume that the bullet flyes 500 m/s and that a 20cm distance would be fine.

By my calculations:

1m/0,2m = 5

500m/s * 5 = 2500m/s

10 times 2500m/s = 25000m/s

1/25000 = 0,00004 sec

0,00004 / 8200ohm = 4,87 * 10^-9

The capasitor then needs to be 4,7nF? Isnt that right?

Lets say I read 2volts on my multimeter and my battery is a 9volt, then the calculations afterwards would be this:

t = (R*C) * (Vc / V)

t = (8200*0,0000000047) * (2 / 9)

t = (0,00004 * 0,2222) = 8,89 * 10^-6

speed = d / t

speed = 0,2m / 8,89 * 10^-6 = 22502m/s

And that is not the result I think is true..

Have I made a mistake or does this simple construction not work.

Last edited: Aug 15, 2009
8. ### beenthere Retired Moderator

Apr 20, 2004
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Subject to controlling the exact amount of charge on the capacitor before the shot, that might work. The component parts are inexpensive enough to permit a bit of experimentation. You could even use a heavy pendulum as the wire-breaker, to see if a consistent result may be arrived at.

On the other hand, for really precise results, the circuit linked to is much better. The down side is the extra time and effort. It is a good investment to make personally if you have a continuing interest in electronics. Having to learn something should not be seen as an inconvenience or waste of time. Everything you learn turns out to be valuable.

Jul 7, 2009
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Except when you're capable of hiding your own Easter eggs...

10. ### Bernard AAC Fanatic!

Aug 7, 2008
4,140
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Might use higher voltage for charging cap, say 18V. Adjust RC & distance [ 100 cm ], to give about a 1V charge at 3000 ft/sec. [914.4 m/s ]. Use a impedance booster for meter, 0 gain noninverting operatoinal amplifier.

11. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
0
Okay thanks for the advice Bernard. I would love to make that circuit you link to Beenthere and I know that it would take me very long time to make it. But at the moment I am more interested in finding the most simple circuit.

12. ### Bernard AAC Fanatic!

Aug 7, 2008
4,140
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The broken wire system is about as simple as I have seen. With accurate components and a good digital voltmeter, with booster, system should work. A .1μF cap, 10kΩ resistor, 3 ft spacing, 9V supply should give about 5V for 3000ft/s [ one msec travel between gates ]. Increase battery V or increase resistor to operate in a more linear region.
Buffer amp is simple. Use a J fet input OP amp, like LF353N, with two 9V batteries to boost input to 10*12 Ω, per Forrest M Mims III. +9V to pin 8.

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13. ### Søren Senior Member

Sep 2, 2006
472
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Yes.

500 m/s is (1/500 =) 2 ms/m
One fifth of that (the 20 cm) is 400 µs

If you wanna use the RC charge time, you'd be wise to opt for the assumed speed to be somewhere around 1/2 to 2/3 in the charge curve.
This means you need an RC-time of ~1 ms and in your case, just go with a tau (R*C) of slightly less, say 800µs to 1ms.
22 kOhm and 47 nF for start values should get you close to something useable.

That said, I'd go for a digital counter and it can be made reasonaably simple, especially if you would be OK with a binary display.

Btw. Do you live anywhere near Copenhagen?

14. ### Jacob J Thread Starter Active Member

Jun 18, 2009
159
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No I am not, I am living in Århus. Do you have some simple schematics for a digital setup?

15. ### Bernard AAC Fanatic!

Aug 7, 2008
4,140
393
Gate wires; 'have used mechanical pencil lead, breaks clean-no stretch, .7mm B , about 2 ohms, 6 cm long. Works with any system.