How bright is the Sun compared to the sky?

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• LightningInAJar
In summary, the brightness of the sun compared to the sky that it lights up is much greater than 19 stops.

LightningInAJar

TL;DR Summary
Sun versus sky brightness?
What is the brightness of the sun compared to the sky that it lights up? I have a solar filter that only sees the sun. I just wonder if it is needed to be so dark that I can't see the sky anymore. I have plenty of camera options left to shorten exposure even with a filter a fraction as dense.

https://www.britannica.com/science/luminosity

"The luminosity of the Sun is 3.846 × 1026 watts (or 3.846 × 1033 ergs per second). Luminosity is an absolute measure of radiant power; that is, its value is independent of an observer’s distance from an object."

Moved...

If the atmosphere is 90% transparent and we assume all of the rest is scattered, I calculate based on the area of the sky and sun that the sun is 400,000 times brighter than the sky(someone should check me). So, no, not feasible to get the sky and the sun in the same photo.

russ_watters said:
I calculate based on the area of the sky and sun that the sun is 400,000 times brighter than the sky
So, about 19 stops? (someone should check me, too!)

That is quite a dynamic range (understatement!)

What do you define as "can't see the sky"? I mean, you can, it's just very dark and uniform.

Do you perhaps mean "can't see cloud patterns in the sky?" In which case, your question would be "how bright are clouds compared..."

russ_watters said:
...
If the atmosphere is 90% transparent and we assume all of the rest is scattered, I calculate based on the area of the sky and sun that the sun is 400,000 times brighter than the sky(someone should check me). So, no, not feasible to get the sky and the sun in the same photo.
I have no idea how to calculate this, but a Journal entry from 1913 states:

... the brightness of the sun is of the order of 100,000 times the brightness of the sky for equal angular areas ...

Since it's difficult for me to determine who is measuring what, I'm going to guess your calculations are correct.

russ_watters
DaveC426913 said:
What do you define as "can't see the sky"? I mean, you can, it's just very dark and uniform.

Do you perhaps mean "can't see cloud patterns in the sky?" In which case, your question would be "how bright are clouds compared..."
I haven't yet attempted to brighten images, but I'm assuming the sky is essentially black and contains no color information in photo.

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russ_watters
OmCheeto said:
I have no idea how to calculate this, but a Journal entry from 1913 states:

Since it's difficult for me to determine who is measuring what, I'm going to guess your calculations are cocorrect.
My assumptions are 90% transmitted, 10% scattered and from there it is a geometry problem comparing "surface areas". Assuming what isn't transmitted is all scattered, not absorbed probably makes up most of my error, but I have no idea what that number is.

Edit: er, wait, that's in the wrong direction. Not sure where the error lied then (10% scattered is too high maybe).

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LightningInAJar said:
I haven't yet attempted to brighten images, but I'm assuming the sky is essentially black and contains no color information in photo.
Yeah, a good depth is 16bit per channel, which is 65,000 levels. Short of the ratio.

LightningInAJar said:
I haven't yet attempted to brighten images, but I'm assuming the sky is essentially black and contains no color information in photo.
Right. So what would you consider an acceptable threshold of visible? The sky alone is essentially featureless. So, clouds?

DaveC426913 said:
Right. So what would you consider an acceptable threshold of visible? The sky alone is essentially featureless. So, clouds?
Yeah, basically clouds. If I did a wide angle timelapse it might be nice to see clouds and sun move and maybe trees too for reference. I wish my camera could take multiple exposure at each interval but I might still need to removal and replace physical solar filter each time then composite each set of images.

russ_watters

Solar radiation that is scattered at least once before it reaches the surface.

As a percentage of the global radiation, diffuse radiation is a minimum, less than 10% of the total, under clear sky conditions and overhead sun. The percentage rises with increasing solar zenith angle and reaches 100% for twilight, overcast, or highly turbid conditions. It is measured by a shadow band pyranometer.

Maybe try with a 4 to 6 f-stop smaller aperture than needed for a landscape exposure and see what you get. This assumes you can do multiple-exposure frames (or post process) with the camera on a tripod.

(above found with:

Let us know what works -- or doesn't.

Cheers,
Tom

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I'm still not sure what the OP is looking for. @LightningInAJar maybe you can say? Do you want a sun that isn't blown out, along with clouds that aren't underexposed?

If that's the goal, I think you will need to take two shots at wildly different exposures, and then combine them.

gmax137 said:
If that's the goal, I think you will need to take two shots at wildly different exposures, and then combine them
Yeah, but:
LightningInAJar said:
Yeah, basically clouds. If I did a wide angle timelapse it might be nice to see clouds and sun move and maybe trees too for reference. I wish my camera could take multiple exposure at each interval but I might still need to removal and replace physical solar filter each time then composite each set of images.

russ_watters and gmax137
I don't suppose they make rotating filter holder that can quickly flip between filters right in front of lens for such a specialized task?

LightningInAJar said:
I don't suppose they make rotating filter holder that can quickly flip between filters right in front of lens for such a specialized task?
What would that be for? Making one exposure? There's no way to remove it fast enough. You could try going the other way and doing a long exposure with the filter on.

I think composite is the way, but even then you have to be careful not to damage the camera if you want the same piece of sky the sun is in. And it still might not even come out, as bright as the sun is. You might try making a sun blocker for the camera (like a pinhead suspended in the frame to cover the disk).

LightningInAJar said:
I don't suppose they make rotating filter holder that can quickly flip between filters right in front of lens for such a specialized task?

When imaging the sun, use the filter (you mentioned that you have and use a solar filter). I advise that you don't risk overheating your sensor and optics by imaging the sun without it.

Instead, here's a way you can get a rough, yet fairly accurate answer to your original question:

(Optional step 0: If your camera/software supports it, set up your camera to record/display images in monochrome [black-and-white]. While this isn't absolutely necessary, it will make measurements easier.)

Step 1: Sun's acquisition:

Image the sun as you normally would with the filter. In this image, the sun's disk should not be saturated. In other words, the detail of the sun should be clear, and not blown out. It's okay of the surrounding sky is dark.

Take note of the exposure time used.

Also take note of the approximate pixel values of the sun's disk (this will be easier if the images are being displayed in black-and-white). If you don't have access to pixel values, look at your histogram. You should see a little blip on histogram corresponding to the sun. The sun will correspond to the right-most part of the curve that's not zero valued. The magnitude of the blip (on the vertical axis) depends on your focal length and sensor size. The larger the sun's disk is relative to the entire frame, the larger the hight of that blip will be. Take note on where in the histogram (on the horizontal axis) that blip is. It should be somewhere roughly in the middle of the histogram, more-or-less.

Step 2: Sky acquisition:

Don't change anything, except for the direction your camera/lens/scope is pointing. Move it away from the sun such that it points to a blank area of sky (such that the sun, or anything else, is not in the frame). Do not change the camera's gain/ISO. Leave the filter on.

The only think you're going to change (besides the direction) is the exposure time. You can expect it to be significantly longer for the sky shots.

Increase the exposure time such that the pixel values of the sky image match those of the sun's disk in the previous image (again, this will be easier if you're working with black-and-white). The histogram for the sky image will be a simple spike. As you increase the exposure time the spike will move to the right on the histogram. Adjust your exposure time such that the spike is near the location (on the horizontal axis) where that blip was on the sun exposure shot.

Take note of the exposure time for the sky shot.

Interpretation:

Divide the exposure time of the sky shot by the exposure time of the sun shot, and that's difference in intensity between the sun and the sky. This method isn't necessarily the most accurate method, but it should give you a good ballpark value.

If you want to convert that to "stops," like those used in conventional photography, you can calculate that using

$\mathrm{stops} = \log_2 \frac{T_{Sky}}{T_{Sun}} = \frac{1}{\ln 2} \ln \frac{T_{Sky}}{T_{Sun}},$

which is approximately

$\mathrm{stops} \approx 1.443 \left( \ln \frac{T_{Sky}}{T_{Sun}} \right).$

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Tom.G