Repainting EF 70-200mm f/2.8L

[eugeniu_sofroni]

WoW ... you guys are something<br><br>

 

I needed help on tips for painting this great lense and now i have a full blown out war on whites vs blacks !. The lens I have my eye on is a "Storm Trooper" and it has been used a quite a lot so there is lots of ware on the body and a lot of the lettering marks on it is missing. <br><br>

 

<img src="http://dosan.skku.ac.kr/~sjkim/icons/Hobby/Figure/StormTrooper(Medicom)_01.jpg"><img src="http://i23.ebayimg.com/03/i/000/ab/4b/139d_1.JPG"><br><br>

 

The lens was R&R-ed and is functioning great so I would like to treat the body the same way the interior and mechanics were treated. I was thinking i could use some varsol to remove the rest of the lettering so it does not look sooo haggared or repaint the beast and if i am going torepaint it i am not going to repaint it the stock color. <br><br>

 

I would rather spend the money to paint it black and buy the white LensCoat and use it when needed, because i will be shooting in the city mainly with low light.

 

For all you geeks here and all these physiscs talk ... you are aware that there are specially formulated heat repeling paints that are all "BLACK", right !<br><br>

 

"<i>Luke, come to the dark side !</i>"<br><img src="http://msnbcmedia.msn.com/j/msnbc/Components/Photos/050411/050411_darthVader_hmed2_3p.hmedium.jpg"><img src="http://images3.speurders.nl/images/31/3100/31004003_1_detail.jpg"><br><br>

 

 

"<i> I see my red door and it has been painted BLACK ...

I wanna see it painted, painted, painted, painted BLACK </i>"

[mark_chappell]

<I> For all you geeks here and all these physiscs talk ... you are aware that there are

specially formulated heat repeling paints that are all "BLACK", right !</i><P>

 

A black paint in sunlight is going to absorb more solar energy, at least in the visible

spectrum, than a white paint. That may not necessarily be the case for other

electromagnetic wavelengths, but to reduce absorptivity of energy, the paint would have to

be reflective (i.e., not 'black' for those wavelengths). <P>

 

It's possible you may be thinking of the black paints sometimes used to increase heat loss

from metal surfaces, but their function is to increase emissivity, not reduce absorptivity

(shiny metal surfaces may have low emissivity and absorptivity in thermal wavelengths).

One example: the Lockheed 'blackbird' aircraft, which were painted black to increase heat

loss and help protect against the high temperatures generated from air friction at Mach 3+

speeds.

[walang_pangalan]

Mr. White, the equilibration times are likely going to be on the order of minutes for a low-mass thing like a lens. But that's just me waving my hands: only a direct measurement can say for sure. Convective effects are going to be much more important in practical use. And, yes, it's unlikely the lens will stop working even if it is asked to perform at 130F.

 

As for Mr. Chappell's conversion argument: he forgets that (a) the solar spectrum, even at the surface of the Earth, is dominated by the near-infrared and (b) all common paints, be they white, black, or even pink, are more or less equivalent absorbers there. If we accept his (reasonable) claim the objects in question at far infrared are also about the same, the result stated again follows: you'll observe the same temperature.

[mark_chappell]

<I>As for Mr. Chappell's conversion argument: he forgets that (a) the solar spectrum, even

at the surface of the Earth, is dominated by the near-infrared and (b) all common paints,

be they white, black, or even pink, are more or less equivalent absorbers there. If we

accept his (reasonable) claim the objects in question at far infrared are also about the

same, the result stated again follows: you'll observe the same temperature.</I><P>

 

Walang, check your facts. Roughly half of the energy in the solar spectrum is contained in

visible wavelengths at the Earth's surface (http://en.wikipedia.org/wiki/Solar_radiation

and other sources), and the difference in integrated solar absorptivity between a 'black'

and a 'white' object can be 50% or more, although it's usually closer to 20-30%. For

example, typical integrated solar absorptivities for dark animals are on the order of

80-90% versus maybe 60-70% for very white animals. White paints are often considerably

more reflective of solar radiation than animal surfaces and it's quite feasible for a good

white paint to exceed 50% reflectance (i.e., less than 50% absorbance).<P>

 

If that doesn't convince you, try this simple experiment, which I (and many other thermal

biologists) have done repeatedly in teaching labs: take two otherwise identical soda cans,

one painted black and one painted white, and place them next to each other in sunlight.

The equilibrium temperature of the black can will be considerably higher than that of the

white can; the difference is highly dependant the size of the cans, orientation, and on wind

conditions. <P>

 

Similar factors will affect the temperature of a white or black lens: how big it is, how it is

aligned with respect to the solar beam, how much of it is really 'white' (versus dark-

colored focus rings and so forth), how reflective the paint is, air temperature, wind, etc.

I'm not sure that for a 70-200/2.8 the average temperature difference between white and

black lenses would be substantial, but if they are in the sun there <B>will</b> be a

difference.<P>

 

If you check my website and explore the publication list, you'll see that I've done quite a

bit of research on thermal balance in animals, so I have some background in this stuff.

[lee_shively]

Any of you physicists ever used a large format camera and worked under a dark cloth in the Arizona desert? Try it sometime. You will quickly understand why dark focus cloths are available with one side made of white material.

[jwhite3.0]

Eugeniu, did you try contacting Canon to see if you can buy touch-up paint? I know from time to time you can find the off-white Canon-paint on Ebay so maybe the black paint will show up too. How about just going to a pro repair shop near your area and ask about buying a little jar of the stuff from them? Electrical tape will work as well. How about a hobby shop or finding some theater prop guys that make stuff who always amaze me with their knowledge of painting metals, wood, plastics, etc?

 

Good luck and thanks to everyone for reminding me why I plugged all the physics equations in my TI-81 (which I still use everyday!).

[eugeniu_sofroni]

I have made up my mind ... if i get this lens it is going black.

 

I just need some input about HOW to paint it.

 

Here is what i have figured so far:

- Take it to Canon Repair Center

- Have them take it appart

- I ask for the metal parts

- I sand blast them, primer and paint

- They reassemble the lens wit the newly painted parts

 

:-)

 

Is there anything stopping me from doing that or messing something up ?

[mark_chappell]

<I>Any of you physicists ever used a large format camera and worked under a dark cloth in

the Arizona desert? Try it sometime. You will quickly understand why dark focus cloths are

available with one side made of white material.</i><P>

 

Never tried this specifically, but there is a pretty obvious difference in comfort level if you

stand in the sun on a hot day wearing a black vs. white shirt.

[des adams]

I wouldn't know but aren't condoms very stretchy ... you can get black ones.

[chris_laudermilk]

Did anybody even read the link Dave put in the first response before running off on tangents? I don't know about you but I won't assume I'm smarter than Canon's engineers and will go with their reason for doing it. At this point, I agree there is a marketing angle to it as well.

 

Personally I wouldn't repaint the lens, especially black. From the picture it appears to be a 70-200/2.8L IS which does have fluorite elements. Again, I'd figure the engineers know more about it than we do.

[alec_myers]

"Thermodynamic argument (and observed reality for that matter) says that, for an object in thermal equilibrium with its surroundings, emissivity == absorptivity. This means that two objects in radiative contact with some reservoir (aka "the Sun"), they will both come to the same temperature, regardless of their color or other properties."

 

That's just plain incorrect, I'm afraid.

 

The objects in question don't reach thermal equilibrium with the temperature of the radation source (the sun) or they'd be at 5700K and have melted. The heat loss by radiation from an object at earth-temperatures is dwarfed by heat loss through conduction to the air. Because black objects absorb more radiation than white ones, they get hotter. Period.

 

Secondly it's only true for black body radiators. While many things are approximations to black body radiators, very few are. Two objects with different finishes can easily have different spectra at the same temperature.

[mark_chappell]

<I>The heat loss by radiation from an object at earth-temperatures is dwarfed by heat

loss through conduction to the air. </i><P>

 

Well.... since we're still beating this to death let me smite the dying carcass once or twice

more.<P>

 

The balance between convection and radiation in heat exchange and temperature

regulation depends on many factors but

mainly air temperature, wind speed and size: the bigger the object is, the greater the

importance of radiant load and the lesser the effects of convection. The closer air

temperature is to object temperature, the smaller the effect of convection. On a hot day

(air temperature close to skin temperature), convection has very little effect, but radiant

loss is substantial and is largely independent of air temperature. In the absence of a

strong radiant source like the sun, <B><I>net</i></b> radiant load may be small

because absorbed radiation is about the same as emitted radiation. <P>

 

Radiant load is quite large in full sunlight: incident solar radiation is roughly 1000 watts

per square meter; absorb 70% of that (typical for a reasonably light-colored surface) and

that's still 700 watts per square meter. Depending on posture a human intercepts maybe

1/3 of a square meter, so that's 230 watts of absorbed sunlight. By comparison one's

resting metabolic rate is only about 100 watts.<P>

 

Anybody who's entered a car that's been parked in the sun for a few hours knows how

strongly sunlight can affect the temperature of a fairly big object.

[tuan_tran]

This is so easy to prove! Take two pieces of metal (aluminum or steel) or plastic (polycarbonate, carbon graphite). Paint one black and paint one white. Leave it out in the sun and take a measurement every 15 minutes. I wonder which will be hotter?

[alec_myers]

"Radiant load is quite large in full sunlight: incident solar radiation is roughly 1000 watts per square meter; absorb 70% of that (typical for a reasonably light-colored surface) and that's still 700 watts per square meter."

 

A quick calculation shows that a surface absorbing and re-emitting radiation at a rate of 700W/m^2 would reach an equilibrium temperature of just over 60 Celsius, in the absence of any other heat loss mechanism. If it's a perfect black-body radiator. Which isn't impossible for an object left out in the sun. So perhaps, yes, convection isn't important.

 

Which leaves me thinking that the undeniable difference in the equilibrium temperature of objects of black and white finishes (I have a black car - I *know* the difference) in the summer sun is because although the absorbtion factors are different in the visible region (white vs. black) the emissivities in the infra-red are similar.

[mark_chappell]

<I> So perhaps, yes, convection isn't important.</i><P>

 

It's not important in 'real life' if fluid (i.e., air in most circumstances) temperature is about the

same as surface temperature. Convection depends utterly on the presence of a temperature

gradient. But as the temperature difference between fluid and surface increases, so does

convection. And obviously, convective exchange in the presence of a temperature gradient

increases as wind speed increases (roughly as the square root of wind speed). And the

smaller an object is, the more it is influenced by convection, i.e., the more tightly 'coupled' it

is to air temperature.

[keith_lubow]

The whole white reflecting more light than black argument is just a side note here. I can see wanting it black as well, but not if I was shooting in the open sun for long periods of time. Well, if you see a white lens and must have it painted black...I still suggest gaffer tape, as in my above post. It has never taken any paint off of anything in my experience, and is very easy to work with and get looking pretty neat.

 

Keith

[eugeniu_sofroni]

the whole reason why i wanted it repainted is so i would not look so haggard if i put tape on it it will look even more haggard ... i want a pleasing / fresh looking lens.

[scott_elowitz]

Here's what the lens looks like with a black LensCoat.

http://www.lenscoat.com/product_info.php?cPath=2&products_id=110

 

or if you prefer "Canon white"

 

http://www.lenscoat.com/product_info.php?cPath=2&products_id=150

[eugeniu_sofroni]

I am ready to close this thread ... i went ahead and bought the older 80-200mm f/2.8L<br><br>

 

Mu ha ha ... i went to the dark side !<br>

 

<img src="http://www.the-reel-mccoy.com/movies/2005/images/StarWarsEpisodeIII_1.jpg">

 

<br>Thanks a lot ppl for the feedback.

[jay_banks]

<p>I am almost ready to switch to Nikon so I don't have to get a white 70-200mm 2.8L for my concert and street photography. Good suggestion for 80-200mm f/2.8L</p>