Off Topic > General Discussion

Photogrammetry help

**Kiwi**:

I took a nice shot of part of New Zealand's South Island from my beach in the North Island back in January, and want to figure out, from a detailed 1:50,000 topographic map, the names, locations (latitude & longitude) and heights of the hilltops in the photo.

To do this, I need to work out the angle subtended by the visible landforms on the horizon, but don't know how to do that, and guess that I'll have to count about 4000 pixels across the width of the original. Can any of our resident experts please help?

From memory, the lowest hilltops at far left and far right are around three degrees apart, but because the land was exceptionally clear (it's most often not visible) and photogenically basking in the summer twilight and looking better than I had seen it in about 50 years, I didn't think to do the usual fingertips at arm's length trick.

The camera is a Panasonic Lumix FZ40 and the zoom lens was set on 54.70mm, the equivalent of 304mm on a 35mm camera.

The detailed EXIF data the camera provides follows:

Filename - 2012-01-17 13.JPG

Make - Panasonic

Model - DMC-FZ40

Orientation - Top left

XResolution - 180

YResolution - 180

ResolutionUnit - Inch

Software - Ver.1.0

DateTime - 2012:01:17 21:06:41

YCbCrPositioning - Co-Sited

ExifOffset - 636

ExposureTime - 1/80 seconds

FNumber - 4.00

ExposureProgram - Normal program

ISOSpeedRatings - 400

ExifVersion - 0230

DateTimeOriginal - 2012:01:17 21:06:41

DateTimeDigitized - 2012:01:17 21:06:41

ComponentsConfiguration - YCbCr

CompressedBitsPerPixel - 4 (bits/pixel)

ExposureBiasValue - 0.00

MaxApertureValue - F 2.83

MeteringMode - Multi-segment

LightSource - Auto

Flash - Flash not fired, compulsory flash mode

FocalLength - 54.70 mm

FlashPixVersion - 0100

ColorSpace - sRGB

ExifImageWidth - 4320

ExifImageHeight - 3240

InteroperabilityOffset - 10798

SensingMethod - One-chip color area sensor

FileSource - DSC - Digital still camera

SceneType - A directly photographed image

CustomRendered - Normal process

ExposureMode - Auto

White Balance - Auto

DigitalZoomRatio - 0.00 x

FocalLengthIn35mmFilm - 304 mm

SceneCaptureType - Standard

GainControl - High gain up

Contrast - Normal

Saturation - Normal

Sharpness - Normal

Maker Note (Vendor): -

Thumbnail: -

Compression - 6 (JPG)

Orientation - Top left

XResolution - 180

YResolution - 180

ResolutionUnit - Inch

JpegIFOffset - 11764

JpegIFByteCount - 2445

YCbCrPositioning - Co-Sited

Another thing I want to figure out is more likely the province of navigation or surveying experts: How much of the hills' heights are showing in the photo?

The highest point is most likely Mount Stokes, 1203 metres high, and theoretically its top is 128.75 km from the camera. However, it is part of a number of ranges, so other hills in the photo might be a few kilometres closer to or further from the camera.

Geographic Range Tabels from nautical books and articles about Earth's horizon indicate that at 129 km distance, anything lower than 1121 metres would be below the horizon when viewed from sea level at my beach. The formula I use, which takes refraction into account, is: Multiply the square root of the height of the object (in metres) by 3.845821 to get its range in kilometres, so turning it around and working from the distance, 128.75km divided by 3.845821 = 33.847789718 and that squared = 1120.77 metres height.

However, none of the information I've seen mentions that most observers would have their eyes above sea level, so how does one factor in the height of the observer? At a rough guess I was around 20 metres above sea level when I took the photo. Again, I forgot to check.

**Chew**:

--- Quote from: Kiwi on March 06, 2012, 07:52:01 AM ---Another thing I want to figure out is more likely the province of navigation or surveying experts: How much of the hills' heights are showing in the photo?

The highest point is most likely Mount Stokes, 1203 metres high, and theoretically its top is 128.75 km from the camera. However, it is part of a number of ranges, so other hills in the photo might be a few kilometres closer to or further from the camera.

Geographic Range Tabels from nautical books and articles about Earth's horizon indicate that at 129 km distance, anything lower than 1121 metres would be below the horizon when viewed from sea level at my beach. The formula I use, which takes refraction into account, is: Multiply the square root of the height of the object (in metres) by 3.845821 to get its range in kilometres, so turning it around and working from the distance, 128.75km divided by 3.845821 = 33.847789718 and that squared = 1120.77 metres height.

However, none of the information I've seen mentions that most observers would have their eyes above sea level, so how does one factor in the height of the observer? At a rough guess I was around 20 metres above sea level when I took the photo. Again, I forgot to check.

--- End quote ---

You need to compute the distance to the horizon for your height of eye then work backwards to find what height of eye would be required to make up the difference in range. If your HoE was 20 m then the distance to the horizon was 17.2 km. Subtract that from the distance to the summit 128.75 km = 111.55 km. A distance to the horizon of 111.55 km would require a HoE of 841.3 m. Subtract that from the height of the mountain 1203 m to get the visible portion = 361.7 m.

P.S. Keep in mind that formula assumes standard refraction. You may have experienced some atmospheric towering.

**Chew**:

One way to determine if towering was taking effect can be found in The American Practical Navigator (35 MB PDF). In the Table of Content click on "Tables" then scroll down to the explanation for Table 15 to see the formula for "Distance by Vertical Angle Measured Between Sea Horizon and Top of Object Beyond Sea Horizon". You have all the variables needed to work it out. If the distances match then you had normal refraction.

I've done this procedure numerous times with a sextant, completely for professional curiosity. You should be able to find the angle from the picture. I've found the distance from my ship to Mount Fuji, a flat top mountain east of San Diego, and a few islands around there.

In the old days sailors would go up and down ladders until the light from a lighthouse went below the horizon. They knew their height of eye at the time and the height of the light so they could determine the distance to the lighthouse. It was called "bobbing a light". I "bobbed" a light on San Clemente Island on my first submarine. I had the Officer of the Deck slowly go deeper until the light went below the horizon, I marked our depth, from the depth you know how high the periscope is sticking out of the water, i.e. your height of eye, so I found the distance to the light.

**Kiwi**:

Many thanks, Chew, some very useful information there. I figured it would work something like that, but my brain isn't as sharp as it was a few decades ago and I couldn't work it out. It's after 1am now so I'll study what you wrote later today and hopefully follow it properly then.

Your figure of visible portion = 361.7 m will help me work out the approximate heights of the two lower hilltops at the sides and therefore possibly identify them on the map. If I can do that I won't even need to figure out the angle they subtend in the photo, although I'd still like to know how to do that for future use.

The 35Mb PDF might be too big as I'm not on broadband, but I'll have a go.

Haven't come across the term towering before. In the photo above and another of a closer island taken at the same time, the bits that meet the sea horizon all bend to almost vertical in the last few pixels. Is that the same thing or something similar?

When I look at Mount Egmont/Taranaki (about 155km away) through binoculars, it's interesting to note the changes in what can be seen near the horizon in different weathers. A large and prominent outcrop on the side is visible just above the horizon at times and not at others.

**Chew**:

--- Quote from: Kiwi on March 07, 2012, 07:17:28 AM ---The 35Mb PDF might be too big as I'm not on broadband, but I'll have a go.

--- End quote ---

Oh, right, I forgot. You're still in the Stone Age down there. Here's a screen cap:

"Corrected vertical angle" means you have to correct for height of eye. That is sqrt(height of eye in meters) * -0.0255°. I know, it's weird. Correct for height of eye twice, once before even using the formula? Yep. If the summit to horizon subtends 0.5° (I'm just guessing, I haven't tried to measure the angle.) and your height of eye was 20 m, then the height of eye correction is sqrt(20)*-0.0255 = -0.114°. Add that to 0.5° to get the corrected vertical angle of 0.386°

--- Quote ---

Haven't come across the term towering before. In the photo above and another of a closer island taken at the same time, the bits that meet the sea horizon all bend to almost vertical in the last few pixels. Is that the same thing or something similar?

--- End quote ---

Towering and looming are related effects. They both both appear to elevate objects higher than normal but towering distorts the object more.

ETA: Re-uploaded the screen cap to fix a math error.

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