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Other WWI Aviation Airfields, equipment, tactics, training, uniforms and all other WWI aviation topics

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Old 10 May 2005, 06:10 PM   #1 (permalink)
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How to read a sundial

I have a WWI photo with a date, but no time... I'd like to know precisely what time of day the photo was taken. The sun was shining brightly and shadows are clear.

There's got to be some way of calculating the sun's angle against the shadows and figuring out what time of day it was at the moment the photo was taken, but I'm not bright enough to figure out what it is.

Can anyone help?
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Old 10 May 2005, 10:15 PM   #2 (permalink)
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Yes, as long as you also know the date and location, it's not difficult to get roughly. Having a fair idea of the orientation helps as well. Accuracy would depend on many factors, such as the angle of the shot, clarity of features and shadows.

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There's got to be some way of calculating the sun's angle against the shadows and figuring out what time of day it was at the moment the photo was taken.
There's nothing new under the sun, Stephen. A guy named Eratosthenes proved the earth was roughly spherical using these techniques about 2500 years ago (he fixed the time and measured the shadows, but it comes to the same thing).

PM me and I'll get my calipers and nautical tables out.
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Old 11 May 2005, 05:05 AM   #3 (permalink)
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Just keep in mind that the British used something called double summer time (like daylight savings time) after 1916. Time references I've seen in British memoirs see kind of off because of that.

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Old 11 May 2005, 07:33 AM   #4 (permalink)
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Thanks, guys. I knew it could be done (by someone who knew something). I'm looking through my pile of photos right now and we may have a selection to choose from, which could help our results.

What's PM? Postal Mail? Or can I email you the photos and then you can print them out?
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Old 11 May 2005, 09:15 AM   #5 (permalink)
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PM = private message.

I'd be interested in Duckman's results.

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Old 12 May 2005, 07:34 PM   #6 (permalink)
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Me too!

The theory is fine, but it remains to be seen how well we can apply it in practice.

Stephen, I've emailed you with my addy. Send me some photos, and we'll see how it goes.

This'll be interesting

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Old 13 May 2005, 11:25 AM   #7 (permalink)
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To get it precise you would also need to know the latitude of the location (a shadow cast at one latitude will not be the same length as a shadow cast at the same moment at a different latitude, the exact height of an object in the photo that is casting a shadow, the exact length of an object in the photo nearby to measure the length of the shadow, and have reasonable expectations that the height of the ground at the tip of the shadow is at the same height as the base of the object casting the shadow and that the lens on the camera used was reasonably flat so that there would not be distortions in lengths from one spot of the photo to another - the length of ground encompassed by say, one inch, at the center of a photo will be slightly greater than the length of ground covered by one inch at the edge of a photo since the curvature of the camera lens "stretches" thinks slightly. An extreme example of such stretching is seen when a highly curved lens is used and you get the very obvious "fish-eye" effect.
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Old 16 May 2005, 08:35 AM   #8 (permalink)
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Those are good points; thanks. I think the photos I have will satisfy most of those requirements... very flat ground, fairly consistent camera lens, the exact latitude is known, etc., etc.

It would seem that the exact height of the object casting the shadow is not necessary, because the actual calculations would have to be made off of a percentage basis anyway... for instance, the shadow is (for example) 74.1% as long as the object which casts it, therefore we can make the proper calculations without knowing the height of the subject.
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Old 16 May 2005, 09:36 AM   #9 (permalink)
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My error - I was thinking you may have been refering to an aerial photo which would have kept you from using the direct percentage of gnomen height to shadow length method since the object casting a shadow (gnomen) in an aerial photo is optically foreshortened. For a ground taken photo you can use the direct percentage technique so long as the shadow is in the same relative plane to the object casting the shadow. To the extent the shadow is either coming towards the camera or going away from it to some degree you would need to account for the foreshortening effect to get as close as possible to an exact answer.
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Old 16 May 2005, 09:39 AM   #10 (permalink)
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My error - I was thinking you may have been refering to an aerial photo which would have kept you from using the direct percentage of gnomen height to shadow length method since the object casting a shadow (gnomen) in an aerial photo is optically foreshortened. For a ground taken photo you can use the direct percentage technique so long as the shadow is in the same relative plane to the object casting the shadow. To the extent the shadow is either coming towards the camera or going away from it to some degree you would need to account for the foreshortening effect to get as close as possible to an exact answer. I don't have a formula for compensating for foreshortening, but I'm sure someone has calculated a trigonometric function for its determination somewhere.
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