High Speed Photography

[Kyle Frasure]

Hello all. My name is Kyle Frasure and I am by no means an expert photographer (as will become more apparent as you read on). I have what seems to be quite an ambitious idea in my head to photograph the explosion of a glass bottle being shot by a gun. I have been doing quite a bit of reading about shooting in high speed and feel like I have a general idea about how to do this but I have questions as to how to perfect this shot.

I understand that synchronizing the flash, shutter, and action is typically the biggest problem that needs to be addressed. My idea is to build a box that would house the bottle. The plane behind the bottle would serve as the background for the photograph (I was thinking a matte black background in order to capture more detail). I am trying to figure out where to put the flash to show the most detail as the bottle is glass and I am afraid that there would be a reflection if the flash was attached to the camera. I am thinking that cutting a hole in the top of the box and placing the flash downwards focusing on the bottle from the top. The flash would be attached to a sound trigger that I have purchased in order to sync the firing of the gun with the flash. Does this sound feasible? Any suggestions on how to better set up the shot? Should I use 2 flashes?

Another question I have is regarding the settings for the camera. Many sources have said that high speed photographs are usually taken in completely dark rooms in order to keep the shutter open for a relatively long period of time (1-2 sec) and, in effect, have the quickness of the flash capture the image. Is this the only way to capture this type of image? I am not sure where I would be able to fire a gun at night as there are local laws restricting shooting after dusk.

Thank you in advanced for ANY comments and/or suggestions.

 

[Asher Kelman]

To light the bullet and broken glass one can have lights triggered by the bang/flash of the shot and add a time delay until you nail it. With a a good flash the duration can be very short. One can rent a single flash for a day and about $50-70 or less. You just need something that's fast. Generally the higher the power of a low cost flash, the longer the flash duration. However, here you only need about 10 Watt Secs or less so naturally, you will be at the fastest flash setting. With slightly more expensive units, the flash duration can be extremely fast.

Where are you allowed to shoot a gun at a home? I'm not sure if it's even allowed during the day, never mind at night. You could use a silencer

The box can be set up with an additional closed large box of black styrofoam taped with black gaffers' tape in front of your setup each totally blocking the front and marked with you target. If the bullet shoots through the styrofoam, in a dim room, essentially all the light will have to come from the flash.

The big thing is to get a some distance between the gun and the target to allow for adjustment in the timing of the flash.


Here's a way one guy does it!

Sensors exactly 2 inches apart calculates bullet speed.

So what trigger have you found that you like?


Asher

 

[Kyle Frasure]

Asher, thank you for the quick response. I'm glad I took your advice on joining the forum; the information is invaluable. As far as the legality of shooting at home, it is illegal in most cities. I am trying to figure this out but there are .22 cal bullets that have a velocity of 710 ft/sec which is too slow to break the sound barrier, therefore, the loud crack of a typical round is avoided and it sounds like a cap gun. With that being said, the logistics are still being worked out

I will be using a Nikon D100. I have a Nikon Speedlight with the capability of a flash duration of 1/25000 sec. Will this flash work or will I need to get a faster one still?

As far as the sound trigger I have procured. I bought the sound trigger with the delay unit from HiViz. I have been unable to give it a try as I need to purchase one more wire to complete the assembly. I will be giving the set up a try this weekend and will report on my findings.

Thank you.

Cheers,
Kyle Frasure

 

[Asher Kelman]

I'm not sure how good your flash delay is but likely there are suggestions with the circuit you bought. What model are you planning to use? in any case if you are too close, the gap you are working with will be hard to fine tune. So you need to go to as long a path as feasible. Just going from 3.5 feet to 21 feet will move the timing space from 0.005 second to 0.02 sec approx.

Asher

 

[Doug Kerr]

Hi, Kyle,

I am trying to figure this out but there are .22 cal bullets that have a velocity of 710 ft/sec . . .I have a Nikon Speedlight with the capability of a flash duration of 1/25000 sec. Will this flash work or will I need to get a faster one still?

Well, with an bullet velocity of 710 ft/sec (8520 in/sec) and a flash duration of 1/25,000 sec (40 µs), the bullet would travel about 0.34 in. during the duration of the flash.

Doesn't sound too promising.

Best regards,

Doug

 

[Asher Kelman]

Hi, Kyle,



Well, with an bullet velocity of 710 ft/sec (8520 in/sec) and a flash duration of 1/25,000 sec (40 µs), the bullet would travel about 0.34 in. during the duration of the flash.

Doesn't sound too promising.

Best regards,

Doug

Well Doug,

Isn't it possible that styrofoam sheets would slow the bullet?

Asher

 

[Doug Kerr]

Hi, Asher,

Well Doug,

Isn't it possible that styrofoam sheets would slow the bullet

For ball ammo, small tanks of ballistic gelatin might be good.

Best regards,

Doug

 

[Asher Kelman]

Hi, Asher,

For ball ammo, small tanks of ballistic gelatin might be good.

Well Doug, for that matter, ballistic gelatin blocks can be used with any bullet.

Kyle, here's the D.I.Y. recipe. It should slow down the bullet as a function of the radius to the 3rd power, I'd guess although I have no clue what the factor would be. It should behave like air resistance formulas in air conditioning pipes or fluid low in tubes. Whatever drag would occur would precisely slow the bullet proportional to the length of gelatin column travelled. First we can increase the mass and go to a 35 magnum bullet and the speed will be about 600 fps. Then if that can be brought down 50-25 ft per second, hopefully there would be still enough force to break the glass and the bullet would be moving then at .6 to .3 mm./sec which is not bad.

Everything depends on the coefficient of "slowing down" for the bullet. Should the bullet be tiny or large to maximize the drag and slowing? I think it should be large and then we pack a large mass at the exit from the gelatin.

Asher

 

[Doug Kerr]

Hi, Asher,

Well Doug, for that matter, ballistic gelatin blocks can be used with any bullet.

Of course, but I understand hp's may open up "a bit" during transit. Might make them unready for their "closeup".

Best regards,

Doug

 

[Doug Kerr]

Hi, Asher,

. . . hopefully there would be still enough force to break the glass and the bullet would be moving then at .6 to .3 mm./sec which is not bad.

Its ballistic properties might be a tad peculiar at 0.002 ft/sec.

Best regards,

Doug

 

[Asher Kelman]

Hi, Asher,


Its ballistic properties might be a tad peculiar at 0.002 ft/sec.

Best regards,

Doug

Hmmm? How did you get that! You must be working with more decades or more brain cells or both! I just reduced the speed of the bullet and used 1/25,000 sec flash to capture a smaller portion of that transit.

Asher

 

[Mike Bailey]

Unless you're really fixed on using a small caliber .22, you might consider a CO2 cartridge or pump pellet/bb gun. Depending on how pumped-up it is, the speed of the pellet or bb is at most half that of a .22 and you could safely and undoubtedly legally set up something in your basement to do this experiment. The shattering results should be just as entertaining and you would have some control over the pellet speed!

Mike

______________________
Mike Bailey
The Elemental Landscape
http://BlueRockPhotography.com

 

[Doug Kerr]

Hi, Asher,

Hmmm? How did you get that! You must be working with more decades or more brain cells or both! I just reduced the speed of the bullet and used 1/25,000 sec flash to capture a smaller portion of that transit.

Well, you said:

. . .and the bullet would be moving then at .6 to .3 mm./sec

So, to take the top end of that range, 0.6 mm/sec is about 0.024 in/sec.

(0.6 mm/sec)/(25.4 mm/in)=0.024 in/sec
​

And that is about 0.002 ft/sec.

(0.024 in/sec)/(12in/ft)=0.002 ft/sec
​

I'm sure you must have meant to say something else.

Perhaps you meant to say an expected motion of the bullet of 0.6 to 0.3 mm during the duration of the flash (1/25,000 sec).

The top end of that range would correspond to a velocity of about 49.2 ft/sec (33.6 mi/hr).

(0.6 mm)/((1/25,000) sec) = 15,000 mm/sec
(15,000 mm/sec)/(25.4 mm/in) = 590.6 in/sec
(590.6 in/sec)/(12 in/ft) = 49.2 ft/sec
(49.2 ft/sec)•((60 mi/hr)/(88 ft/sec))= 33.6 mi/hr
​

Or perhaps you meant something else entirely.

Best regards,

Doug

 

[Bart_van_der_Wolf]

Well, with an bullet velocity of 710 ft/sec (8520 in/sec) and a flash duration of 1/25,000 sec (40 µs), the bullet would travel about 0.34 in. during the duration of the flash.

Doesn't sound too promising.

Unless Kyle wants to record the exploding bottle, but not necessarily the bullet itself (although that would be cool ...).

One will need to figure out the exact duration of the flash in order to determine if it has any possibility to 'freeze' the motion.

Back in the 80's when I was, amongst others, involved in high speed photography everything was of course film based. We then had a few variables to play with, one of which was rapid sequencing of the exposures to capture the bullet in flight at different positions during transition/penetration. For that purpose extremely fast (turbine driven) cinematography equipment was used and film on very tough tear resistent polyester base was needed to survive the acceleration.

Due to the very short exposure times per frame, the scene had to be lit with very bright/concentrated lights in various settings depending on what had to be recorded.

One type of recording was to image the density/compression of air from shockwaves or turbulence from heat exchange, so-called schlieren photography:

Posted here under the fair use principle, click on the images to see more examples.​

Mind you, as descibed in the Scientific American articles copied on the pageswhere these imaes came from, a light flash discharge was used with 20,000 KiloWatt for 0.3 microsecond output to achieve the exposure on the photographic emulsions of those days. One can also imagine the trigger timing requirements.

For long sequences of images, such as cinematography, a high power continuous lightsource is needed, although it could take mere seconds to devour a 100 feet roll of film. Exciting stuff.

So for Kyle's experiment one has to set an achievable goal, the shattering of glas (or other substances) or the bullet itself, in a single well timed frame. Both will require different levels of light and exposure times, potentially a combined flash and longer exposure in an otherwise darkish setting. Depending on the speed of the bullet, and the allowed displacement of the projected image on the sensor (at a given magnification factor) one will need to calculate the required power during the exposure time interval for an adequate exposure.

Also, given the flash unit intended to be used for the experiment, one needs to find out how it can be tricked into delivering the shortest possible flash at the peak of its emission curve. I don't know the flash in question, but perhaps it has a thyristor kind of shut-off mechanism, if so it should see some reflective surface to force its shutdown. Without additional means to steer the pre-flash X-synchronization, such as offered by a e.g. PocketWizard FlexTT5, we have to assume the on camera flash timing will maximize exposure during the discharge. The light of the flash could be concentrated on the subject to maximize the amount of light. One could think of constructing a snout or using a condensor lens.

Once the technicalities are determined, one can also look at the aesthetics. The positioning of the light will determine how the glass will look, and one can generally choose between a bright field or a dark field lighting setup. Simply put, a bright field lighting consists of lighting the background of the glass, but provide dark edges in the background just outside the edges of the glass or the field of view. Dark field lghting requires the opposite, a dark background and backlighting just outside the field of view, with no other light falling on the glass from the front. For details I recommend reading "Light, Science & Magic".

Cheers,
Bart

 

[Bart_van_der_Wolf]

Perhaps you meant to say an expected motion of the bullet of 0.6 to 0.3 mm during the duration of the flash (1/25,000 sec).

The top end of that range would correspond to a velocity of about 49.2 ft/sec (33.6 mi/hr).

Still assuming the bullet needs to be stopped in motion, don't forget that whatever assumptions are made, the motion on the sensor is reduced by the image's magnification factor.

Cheers,
Bart

 

[Asher Kelman]

Perhaps you meant to say an expected motion of the bullet of 0.6 to 0.3 mm during the duration of the flash (1/25,000 sec).

The top end of that range would correspond to a velocity of about 49.2 ft/sec (33.6 mi/hr).

I'm assuming we can slow down the bullet 10 fold using the ballistic gelatin and then look at the speed of the flash to determine how far the bullet would travel in the interval.

Of course, the flash does not slow the speed just samples a minute duration of the bullets flight.

Still assuming the bullet needs to be stopped in motion, don't forget that whatever assumptions are made, the motion on the sensor is reduced by the image's magnification factor.

Cheers,
Bart

So we'd sop better then with a Nikon with a 1.6 multiplier or for example a digicam like the G10 working with a Canon flash.

Asher

 

[Doug Kerr]

Hi, Bart,

Still assuming the bullet needs to be stopped in motion, don't forget that whatever assumptions are made, the motion on the sensor is reduced by the image's magnification factor.

True, but irrelevant. The size of the bullet , or any other scene object we wish to use as a baseline for evaluating blurring,
is affected the same way.

Any reasonable criterion for "acceptable blurring" will not be affected by format size. If we decide it is acceptable for the bullet to smear by 0.05 times its length, then a setup that yields that will do so regardless of format size.

Best regards,

Doug

 

[Doug Kerr]

Hi, Bart,

Thanks for all this fascinating info.

a light flash discharge was used with 20,000 KiloWatt for 0.3 microsecond output . . .

That comes to 6 watt-seconds! Doesn't sound right, somehow.

(I do see those numbers cited in Stong's Scientific American column speaking of Vandiver's work.)

I'll do a little more poking around on that.

Thanks again.

Best regards,

Doug

 

[Bart_van_der_Wolf]

That comes to 6 watt-seconds! Doesn't sound right, somehow.

20,000,000 W during 0.0003 seconds, if it where possible to maintain that for 1 second, wouldn't that amount to 6.67 x 10^10 W/s? Disclaimer, I'm not an electro-engineer.

Cheers,
Bart

 

[Doug Kerr]

Well, in fact, the electrical details in the article show that the energy storage in the flash capacitor is about 8 joules (8 watt-seconds).

So that is quite consistent with the instantaneous power and the pulse duration cited.

Best regards,

Doug

 

[Doug Kerr]

Hi, Bart,

20,000,000 W during 0.0003 seconds, if it where possible to maintain that for 1 second, wouldn't that amount to 6.67 x 10^10 W/s? Disclaimer, I'm not an electro-engineer.

The time was 0.3 microseconds (0.0000003 seconds). You apparently were thinking of 0.3 milliseconds (0.0003 s).

Energy is the product of power and time (which is why the unit is watt-second). I think you divided (and with the wrong value for time by a factor of 10^3):

(2 x 10^7 W)/(3 x 10^-4 s)=6.67 x 10^10 W/s

W/s is not a meaningful unit here; the wrong unit is the clue that the calculation is upside down!

The correct calculation (with the correct values) is:

(2 x 10^7 W) x (0.3 x 10^-6 s) = 0.6 x 10^1 W-s = 6 W-s.

Best regards,

Doug

 

[Bart_van_der_Wolf]

Hi, Bart,


The time was 0.3 microseconds (0.0000003 seconds). You apparently were thinking of 0.3 milliseconds (0.0003 s).

Energy is the product of power and time (which is why the unit is watt-second). I think you divided (and with the wrong value for time by a factor of 10^3):

(2 x 10^7 W)/(3 x 10^-4 s)=6.67 x 10^10 W/s

W/s is not a meaningful unit here; the wrong unit is the clue that the calculation is upside down!

The correct calculation (with the correct values) is:

(2 x 10^7 W) x (0.3 x 10^-6 s) = 0.6 x 10^1 W-s = 6 W-s.

Best regards,

Doug

Hi Doug,

Thanks for the correction of my 3 zero's (mili/micro) oversight (but then what's 3 zero's, ).

The calculation I follow, but it seems counter intuitive to me. Anyway, the important part is that it resulted in a very brief exposure time, in order to stop motion. Apparently it was bright enough to result in an adequate exposure for the bullet, leaving the shutter open longer than 0.3 microseconds would allow to capture the flame.

Cheers,
Bart

 

[Doug Kerr]

Hi, Bart,

The calculation I follow, but it seems counter intuitive to me.

Well, if the power flow is 2 kW, and lasts for 3 h, the energy transferred is 6 kW-h.

If the power is 3 W, and it flows for 4 s, the energy transferred is 12 w-s.

If the power is 1Mw, and the duration of flow is 1 µs, the energy transferred is 1 W-s (1 joule).

I'm not sure why this should seem counter-intuitive.

Perhaps you were not expecting those quantities to be the ones cited in a photo-flash context . Often we will be expecting one thing, will read something different (but correct), and say, "that doesn't seem right".

Almost never in photoflash contexts do we hear of the power level during the discharge.

So perhaps a calculation involving that sounds "out of kilter".

Often we will hear the total amount of energy stored ("1500 W-s") and a flash duration ("2 ms").

If we calculate the average power (assuming 80% of the stored energy is actually used by the tube before it extinguishes), we are startled to find that the average power comes out 600 kW!

Best regards,

Doug