Algernon Loc: Milwaukee, Wisconsin

Which of these two settings will let in more light, and how much more? (assume identical ISO)
a) f3.5 at 30 seconds
b) f1.8 at 20 seconds

Is there a website that provides a method to do this calculation? My rudimentary online searching came up with some info regarding EV (exposure value) but failed to include my lengthy exposures, and didn't explain how different EV values related to each other.

For background, I am experimenting with Milky Way photos. I am using an NEX-6 (APS-C). I first tried my kit lens at 18mm f3.5. Based on the results I think I need more light capture capability but 30 seconds is the longest exposure acceptable for this mm length to avoid a star "trail".

I also have a prime f1.8 35mm lens. The longer mm length means I can only expose for about 20 seconds to avoid star "trails".

Hence my question. Clear nights with no moon are rare this time of year in Wisconsin, much less having to drive a significant distance to avoid the worst of the light pollution. Therefore I want to be prepared.

I know, I know. The perfect lens would be a wide-angle prime. But I don't have one.

emmons267 Loc: Arizona, Valley of the Sun

You can start here: https://www.flickr.com/photos/samfeinstein/2167243553/

Search 'Equivalent Exposure' on Google, or, if you're really daring - http://en.wikipedia.org/wiki/Exposure_value.

Enjoy - Happy trails.

Algernon Loc: Milwaukee, Wisconsin

Heh, heh. "Happy trails." Well done.

The Wikipedia article is exactly what I was looking for. There is a table there that covers my long exposure times.

Bottom line is that f1.8 for 20 seconds will let in twice the light as f3.5 at 30 seconds.

Thanks!

emmons267 Loc: Arizona, Valley of the Sun

Glad I could help.

NikonDad Loc: Bothell, WA

I think I got this all correct. Some of you smarter folks can step up if need be.

EV = log_2 (N^2 / t ) where:
log_2 is log base 2
N = f number
t = time

You find a log calculator here: http://logbase2.blogspot.com/2008/08/log-calculator.html

a) EV = Log_2(3.5^2 / 30) = -1.290415
b) EV = Log_2(1.8^2 / 20) = -2.625934

b is capturing more light, about 2/3 of a stop more.

If you want a short, rough estimate, you could do this (numbers rounded):
3.5 ---> 1.8 = 1 1/3 stops
30" ---> 20" = -2/3 stop
Difference = 2/3 stop

Algernon Loc: Milwaukee, Wisconsin

Awesome. Thanks for the link. It is more precise than plotting my settings on the Wikipedia graph.

But for the calculation, isn't the difference between the settings 1.33 EV? (not 0.67; 2/3)

NikonDad Loc: Bothell, WA

Hmmmm, yep, you're one of the smarter ones for sure.

I had my eye on the rough calculation and I calculated the number of 1/3 stops incorrectly. f/3.5 ---> f/1.8 is 2 stops. That would produce 1 1/3 stops difference.

Good catch. Thanks.

Algernon Loc: Milwaukee, Wisconsin

Haha. I wouldn't be so sure.

I just noticed your location. Way back in the early 2000s I traveled to Bothell about once a month. Nice area.

NikonDad Loc: Bothell, WA

Well, I'm about as far north as you can get and still be called Bothell.
There's one more house and then you're in Everett. I'm between Lynnwood and Mill Creek. I'm a Bothellite in name only and happy to be so. Don't want anything to do with King County (Seattle). Mary Burke wouldn't have any trouble getting elected here with Seattle's constituency. But, I digress from our topic, sorry.

Algernon Loc: Milwaukee, Wisconsin

To expand on the digression. :)

Next year is my 35th wedding anniversary and I tentatively plan on heading out to your neck of the woods to go on a whale watching trip or two.

We've gone on whale watching trips in Alaska, Hawaii, New Zealand, and the Caribbean. Never have seen a whale. :(

NikonDad Loc: Bothell, WA

To continue the digression (sorry folks), I've lived here 28 years and have never done the whale watching thing.
What time of year will you be here?
You going out of Seattle, the WA coast or Oregon?

Algernon Loc: Milwaukee, Wisconsin

NikonDad, the moderators are going to ask us to take this offline soon. :)

It would be in September. I'd fly into Seattle. My thought was to stay in or near Olympic National Park.

ralphc4176 Loc: Conyers, GA

f/1.8 at 20 seconds will admit more light. If that's not enough light, try shooting at a higher ISO equivalent. You may not have enough equipment to get the exposure you want, but it never hurts to try!

Algernon Loc: Milwaukee, Wisconsin

Thanks ralphc4176. I'm still experimenting*, and shooting at ISO 3200 then again at 6400. I think with the switch to f1.8 the 3200 will work well, but I'll still do both.


* I will be in Patagonia and Antarctica this December and I want to be confident in my selection of settings to obtain the best result possible. I won't get a second chance.

tomw

I would have thought, since the f numbers are reciprocals and the amount of light is relative to their squares, the way to calculate would be:
((1/1.8)^2*20)/((1/3.5)^2*30)
Which gave me a light intensity per unit area of the sensor of about 2.5 times higher for f1.8 for 20 seconds
Am I wrong? I’ve never tried to calculate this before.