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Micromanagement


Jay Young

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20 minutes ago, Satsuki Murashige said:

That would be very Dedo-like! 

Would the fact that Classic 150w Dedos are either 12v or 24v have anything to do with it? The lower voltage was supposed to make the bulbs more efficient.

I can't possibly see how. The reason for higher voltage, up to a point, is to achieve more watts with less current. 150W bulbs at 12V will draw over 12 amps, which is enough to seriously stress XLR connectors and all but the stoutest cables. Double the voltage and halve the current, hence the purpose of high voltage cross-country power lines. I'd assume that's all it was about.

There does come a point where sufficiently high voltages - mains voltage - risk arcing across inside the bulb, given that the pressure in there is typically very low, especially when they're cold. The name Source Four comes from the design of HPL (high-performance lamp) which bundles four filaments into the smallest possible area, with the idea that a concentrated source emits more precisely into an ideal reflector and collimating assembly. It works, but the limits are principally heat, which is why a HPL has a huge heatsink around the bottom, and internal arcing.

But there's no real risk of arcing at such low voltages. More efficient at 12 than 24? I can't see how anyone would substantiate that claim. The opposite would seem to be true.

P

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14 hours ago, Phil Rhodes said:

Isn't that sort of what you'd expect, assuming the behaviour of the control is intended to work in f-stop linearity? 25% to 75% is a 50% increase, 75% to 100% is a 25% increase. 25 is half of 50.

This actually occurred to me shortly after I wrote the question.  I can barely do my son's third-grade math, so... But wouldn't that mean if Dedo's had a 1/4 power (which they don't), then 1/4 power to half power would be two stops, right?  They should make that. ?

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12 minutes ago, Justin Hayward said:

1/4 power to half power would be two stops, right?  They should make that. ?

Yes, but you're talking about not very much actual light at that point.

Going from four photons a second to one photon a second would also be two stops, but both are effectively pitch black!

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7 minutes ago, Phil Rhodes said:

Yes, but you're talking about not very much actual light at that point.

Going from four photons a second to one photon a second would also be two stops, but both are effectively pitch black!

Good point, but I will say a half power dedo is pretty bright these days in the era of 3200ASA.  Maybe there's use for a 1/4 power?  ?

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I thought Dedo said something once about running 150w bulbs at low voltages to generate less heat and thus more lumens per watt, compared to a something like a 120v 1K fresnel. But I might be misremembering. Of course, the Dedolight optics make a huge difference in efficiency as well.

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On 10/26/2020 at 5:36 PM, David Mullen ASC said:

There are a few experienced people who like doing the occasional "get-your-hands-dirty / it's-like-film-school-all-over-again" job. But "occasional" is the operative word.

I have a few fond memories of seeing the director, dp, producer, and tiny grip crew they hired all heave-hoing a big steel deck platform into place on re-shoots after the money's basically run out of a large budget production. You just sit back and appreciate the film-school "everyone's in this" attitude. I think it would go a long ways towards humbling some and building due respect for others when everyone decides to get dirty once in a while. You know, throwing union rules to the wind for an afternoon. This also has the effect of making you appreciate the well oiled machinery of a very efficient and organized production at the same time...

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On 10/27/2020 at 11:54 AM, Phil Rhodes said:

More efficient at 12 than 24? I can't see how anyone would substantiate that claim. The opposite would seem to be true.

Here's a longer explanation than this needs:

With all incandescent lights, which include quartz halogen, the filament is just a resistor that turns electricity into heat. So, you'd want something that burns the hottest before it melts. Of all the elemental metals, tungsten has the highest melting point at about 3700K. (In fact we can't practically make a container to hold molten tungsten, so anything made out of tungsten is sintered together).

Since a filament is just a resistor, the shape of that filament determines how much electricity it gobbles up. Wattage is mainly determined by thickness and voltage is mainly determined by length. As thickness increases, so does wattage and as length increases, so does voltage.

So, a low voltage bulb has a filament that is smaller in length than a higher voltage bulb. A filament is not just a straight line, though, but a coiled coil, so a low voltage filament bundle is also more compact.

Compact filaments structures help in two main ways. One is that because the area of the filament structure is smaller, you can design optics and reflectors that are more precise. You see this precision in Dedolights and MR-16 bulbs.

The other, is that smaller filaments do not need as much electricity to keep them hot (according to this site: https://www.lrc.rpi.edu/programs/nlpip/lightinganswers/mr16/performance.asp ).

With tungsten, the hotter it gets, the more efficient it gets. Technically, tungsten is almost perfectly efficient - it converts electricity to radiation at an almost perfect 1 to 1 ratio. Unfortunately, most of that radiation is outside of visible light - it's in the infrared spectrum.

The hotter you heat tungsten, though, the more that radiation shifts to visible light (otherwise known as lumens).

At 2100K, it's about 4 lumens per watt. 2500K is about 9 lm/w. 3000K is about 21 lm/w. 3200K is about 27 lm/w. 3400K is about 33 lm/w. The limit before Tungsten melts is about 37 lm/w. 

So, the short answer is that lower voltage bulbs are more efficient because they have smaller filaments that more easily burn hotter, which provide more visible light per watt. But wasn't the long answer more fun?

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19 hours ago, Joshua Cadmium said:

Here's a longer explanation than this needs:

With all incandescent lights, which include quartz halogen, the filament is just a resistor that turns electricity into heat. So, you'd want something that burns the hottest before it melts. Of all the elemental metals, tungsten has the highest melting point at about 3700K. (In fact we can't practically make a container to hold molten tungsten, so anything made out of tungsten is sintered together).

Since a filament is just a resistor, the shape of that filament determines how much electricity it gobbles up. Wattage is mainly determined by thickness and voltage is mainly determined by length. As thickness increases, so does wattage and as length increases, so does voltage.

So, a low voltage bulb has a filament that is smaller in length than a higher voltage bulb. A filament is not just a straight line, though, but a coiled coil, so a low voltage filament bundle is also more compact.

Compact filaments structures help in two main ways. One is that because the area of the filament structure is smaller, you can design optics and reflectors that are more precise. You see this precision in Dedolights and MR-16 bulbs.

The other, is that smaller filaments do not need as much electricity to keep them hot (according to this site: https://www.lrc.rpi.edu/programs/nlpip/lightinganswers/mr16/performance.asp ).

With tungsten, the hotter it gets, the more efficient it gets. Technically, tungsten is almost perfectly efficient - it converts electricity to radiation at an almost perfect 1 to 1 ratio. Unfortunately, most of that radiation is outside of visible light - it's in the infrared spectrum.

The hotter you heat tungsten, though, the more that radiation shifts to visible light (otherwise known as lumens).

At 2100K, it's about 4 lumens per watt. 2500K is about 9 lm/w. 3000K is about 21 lm/w. 3200K is about 27 lm/w. 3400K is about 33 lm/w. The limit before Tungsten melts is about 37 lm/w. 

So, the short answer is that lower voltage bulbs are more efficient because they have smaller filaments that more easily burn hotter, which provide more visible light per watt. But wasn't the long answer more fun?


Joshua, thank you for the technical breakdown. Fascinating stuff!

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On 10/25/2020 at 5:58 AM, Satsuki Murashige said:

P.S. I agree with you about the Jimmy Jib - it’s an incredibly flexible piece of gear and when you have an expert operator, there’s not a lot you can’t do with it. And it’s usually just one person, sometimes without even a focus puller. 

Tell that to the 6-man Jimmy Jib crew I had in India, who couldn't (for love or money) manage a basic tracking shot part way down a staircase... or even the simple boom up that I then simplified the shot down to ?

There's some pretty incredible BTS footage of me on the brink of a complete mental breakdown, somewhere around take 38.

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44 minutes ago, Phil Rhodes said:

Joshua, are you conflating wattage and current there?

Certainly the issue of compactness is key, hence the Source Four HPL.

P

If by current, you mean amperage, I'm almost positive you are correct in assuming that at a certain wattage, you have to increase voltage in order for the amperage not to be too high.

However, the main issue with tungsten efficiency is filament temperature and life. A burning tungsten filament sloughs off atomized tungsten that either blackens the bulb (regular incandescent) or is redeposited on the filament due to the halogen cycle in tungsten halogen bulbs (but it is not always redeposited where you want, so parts of the filament still get thinner.) The failure point of a bulb is normally when the filament gets a weak point that breaks - and it increasingly does so at higher temperatures.

While wattage is related to the thickness of the filament, I have read that lower voltage filaments are in general thicker than their higher voltage counterparts (but I have not been able to confirm that). I haven't fully explored the math behind everything, because it gets complicated, but I'm fairly confident that the main reason that lower voltage bulbs are more efficient is that they can be pushed to a higher temperature without the filament breaking down too fast, either because it is thicker, or, since the length is shorter, there is less surface area for weak points to develop. Manufactures test their bulbs by how many fail in a certain timeframe and at some point, the economics make it so that bulbs need to last for a reasonable amount of time. Most of the high efficiency bulbs I've seen burn at 3400K and last for 50 hours.

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Compactness does help with optics, but voltage definitely plays a role in the amount of lumens that can get to those optics, and it is actually apparent in Source Four HPL bulbs. As the voltage goes up, the lumens go down, but the hours stay the same. This is from ETC's data:

750w 240v 300 hour HPL = 19,750 lumens at 3200K  /  26.3 l/pw

750w 120v 300 hour HPL = 21,900 lumens at 3250K  /  29.2 l/pw

750w 77v 300 hour HPL = 22,950 lumens at 3250K  /  30.6 l/pw

So, a 750 watt 77 volt HPL is about 16% more efficient than a 240 volt HPL.

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I have actually been looking at this for awhile because I was trying to determine the highest amount of tungsten light I could run on house power. There is no 50 hour HPL bulb - 300 hours is the shortest anyone makes. (I'm pretty sure it's because no theater is going to want to change bulbs that often.)

So, I started looking at overvolting with a variac, which increases the temperature of tungsten, and thus the efficiency. The 77 volt bulbs (which are normally used on ETC's dimmers) are already more efficient, but they might take overvolting better if the filament is more robust than mains power bulbs. 

If I ran a 77 volt HPL at about 88 volts, it would hit that 50 hour criteria, which should also get me to about 37 lumens per watt. With a 550w 77v bulb, I'd be hitting about 25560 lumens at 680 watts. (Overvolting 750w bulbs might generate way too much heat to be safe). Also, you shouldn't run two 750w bulbs on 15 amp mains power (1440 watts is the 120v 15a continuous limit) so this would help me get close to that limit without going over. It would end up being 17% more efficient than a 750w 120v bulb, but at a lower wattage (or 28% more efficient total). 

On top of that, the efficiencies from overvolting increases when you consider that sensors want to be at 5000K. So, getting closer to that temperature is either going to make the image look better (less saturated red channel) or make it so you lose less light when converting tungsten into 5000K by using gels or dichroics. The difference between 3250K and 3450K is 1/8 CTB, which would be a 25% light loss for gels and a 12.5% light loss on dichroics. 

Now, I don't know how safe overvolting might be on Source Fours. At 88 volts that bulb is going to be around 3500K, which should be okay. There is brief inrush for tungsten, which can take it past the melting point, but you can mitigate that by gently bringing up the voltage from 0. When I talked about overvolting in general with ETC, they said I just risked catastrophic bulb failure (glass shattering). I'm going to be experimenting with this soon, though.

Again, probably more info than you wanted, but perhaps helpful. 

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16 hours ago, Mark Kenfield said:

Tell that to the 6-man Jimmy Jib crew I had in India, who couldn't (for love or money) manage a basic tracking shot part way down a staircase... or even the simple boom up that I then simplified the shot down to ?

There's some pretty incredible BTS footage of me on the brink of a complete mental breakdown, somewhere around take 38.

Lol, well I did say it required an ‘expert operator’! ?
 

But even a decent one should be able to pull off that shot. I’m surprised you didn’t grab the handles and give it a go!

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12 hours ago, Joshua Cadmium said:

While wattage is related to the thickness of the filament, I have read that lower voltage filaments are in general thicker than their higher voltage counterparts (but I have not been able to confirm that). I haven't fully explored the math behind everything, because it gets complicated, but I'm fairly confident that the main reason that lower voltage bulbs are more efficient is that they can be pushed to a higher temperature without the filament breaking down too fast, either because it is thicker, or, since the length is shorter, there is less surface area for weak points to develop. Manufactures test their bulbs by how many fail in a certain timeframe and at some point, the economics make it so that bulbs need to last for a reasonable amount of time.


I’ve owned my Classic Dedos for over 8 years and have never had to replace the globes. That is one of the advertised benefits of Dedos.

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Well, hang on a second. Wattage isn't directly related to the thickness of the wire used to create a filament. A thinner wire will have higher resistance per unit length, but the total wattage will still be a function of both its thickness and its length. Are we confusing two concepts here?

P

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4 hours ago, Phil Rhodes said:

Well, hang on a second. Wattage isn't directly related to the thickness of the wire used to create a filament. A thinner wire will have higher resistance per unit length, but the total wattage will still be a function of both its thickness and its length. Are we confusing two concepts here?

P

This document from Osram is my favorite in terms of explaining how tungsten bulbs work: https://www.bulbconnection.com/file_assoc/63_FL_114103.pdf

It states on page 9 that: "The wire length is mainly determined by the lamp operating voltage. The higher this is, the longer the piece of wire the lamp bulb must contain. The wire in a 24 volt lamp is between 10 and 14 cm long, the wire in a 12 volt lamp is only half that length. The wire diameter is mainly determined by the required lamp wattage and life. The higher the wattage, the thicker the wire, and hence also the stronger it is mechanically."

That seems quite clear cut. I don't know why it is what it is in terms of Ohm's law, just that it is what's described here. I'd love to see the math behind it all, though. 

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Keeping wattage constant, a bulb that operates at a higher kelvin temperature will definitely produce more lumens than a bulb operating at a lower kelvin temperature. But, higher kelvin temperatures will definitely reduce bulb life. Plug in any given bulb here: https://www.ushio.com/support/halogen-lamp-life-estimator/ . You can see the impact temperature will have on lamp life and lumens. 

With classic Dedolights, for instance, one of the reasons the bulbs last so long is because they were designed to operate at 3400K but can operate at 3200K with a drastically increased lifespan. As well, the filament is more robust and will survive mechanical failure from fixture movement much better than line voltage bulbs.

 

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1 hour ago, Joshua Cadmium said:

That seems quite clear cut. I don't know why it is what it is in terms of Ohm's law, just that it is what's described here. I'd love to see the math behind it all, though. 

All they're saying there is that for a wire of a given length making it thicker, which will reduce the resistance, will increase wattage. That doesn't imply that a lamp bulb of a given wattage always has a particular thickness of wire in its filament. It'll  vary with voltage.

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1 hour ago, Phil Rhodes said:

All they're saying there is that for a wire of a given length making it thicker, which will reduce the resistance, will increase wattage. That doesn't imply that a lamp bulb of a given wattage always has a particular thickness of wire in its filament. It'll  vary with voltage.

Oh, I wasn't trying to imply that a bulb of a certain wattage always has a particular thickness, just that higher wattage filaments are thicker than lower wattage filaments all other things held constant. Thicker wires will produce higher wattages, but you have to hold the other part constant, the length of the wire, which is voltage.

I don't know how accurate it is in terms of Ohm's law, but in my mind I picture a certain length of wire that produces a certain wattage. I can stretch out that wire to make it thinner (more volts) or I can squash it together to make it thicker (less volts), but the wattage stays the same.

The main point, though, is that lower voltage bulbs of the same wattage would be more robust - smaller in length and thicker - which I think you agree with as well. (And thicker and shorter filaments survive higher temperatures better, which leads to increased lumens at the same wattage.)

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7 hours ago, Joshua Cadmium said:

That seems quite clear cut. I don't know why it is what it is in terms of Ohm's law, just that it is what's described here. I'd love to see the math behind it all, though. 

Lamp power is P = U*I = U^2/R. 

Resistance of a wire is R = 4*r*L/(pi*D), where L is length, D is diameter. Resistivity r (in Ohm-meters) depends not only on the alloy, but on the temperature as well, which complicates things a lot.

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