What focal length or zoom do I need to photograph a 1-inch railcar seal from up to 96 feet away?

Asked 5/31/2017

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I need to photograph railcar seals from the top of a building to verify that the cap is correct and the steel tie/cable is intact. The longest shooting distance would be about 96 feet, though many shots would be closer to 30 feet. The seal/tag is about 1 1/16 inch long. Reading the engraved text would be a bonus, but the main goal is confirming whether the seal and cable are intact.

Would a bridge or superzoom camera be suitable for this, or do I need something with a much longer lens? I’ve seen cameras advertised as 24x or 60x zoom, but I’m not sure what that means in practice. The photos would be taken outdoors in daylight, and the person using the camera is an amateur, so simple point-and-shoot operation would help. Is there a practical way to estimate the focal length needed?

Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0

Photography Stack Exchange contributor

9y ago

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As the other answers do a good job of giving you the fish, seeing as you're an engineer, this answer will teach you to fish.

The angle of view of a lens & camera combination is simply determined by the isosceles triangle whose base is the diagonal of the camera's sensor size, and whose height is the lens's actual focal length. The vertex angle of the isosceles is the angle of view.

We could actually calculate the desired angle of view using trigonometry, but let's just do some calculation based on similar triangles. Assuming the image you included is roughly the size you need to be able to read the tags, I measure the tag at about 1/12 the width of the image. So the image is 12 3/4" wide. Let's truncate to 12", or 1 ft. Then your subject distance:width ratio (D/W) is 96:1 (in your stated worst case). Using the distance as the height of a isosceles triangle, and the subject width as the base of the triangle, we have put bounds on the angle of view.

In the image below, the gray box represents the camera. D is the distance to the subject (i.e., 96 ft), and W is the width of area to cover (12 inches). Because the triangles are similar, ƒ/d = D/W.

enter image description here

Therefore, by similar triangles, your camera's focal length–to–sensor diagonal ratio ƒ/d should meet or exceed 96:1.

AJ Henderson's answer suggests a Nikon P900 superzoom, which has a "35mm-equivalent" focal length of 2000mm. The sensor size of a 35mm format sensor is 24mm × 36mm, so the diagonal is ~43mm. This gives a ƒ/d ratio of about 46.5:1, or around half of what you would need to produce a similarly size image as you provided, at 96 ft.

Note: the P900 doesn't have a 35mm-format sensor (it's actually 6.2mm × 4.6mm (7.7 mm diagonal)), but since the focal length / angle of view is given relative to a 35mm sensor size in this discussion, we use a 35mm sensor as the calculation baseline. The actual physical max focal length is 357mm (which is 2000/357 = 5.6 times smaller than its "35mm equivalent" focal length). But since the camera's sensor is 5.6 times smaller than a 35mm-format sensor, the scale factors cancel.


So, let's step back. The best focal-length–to–sensor-diagonal ratio you can expect with the P900 is 46.5:1. Is that enough for you? That translates to an image that covers about twice the width of the image you provided, at 96 ft. Put another way, from 96 ft away, a P900 at full zoom will produce an image where the seal tags will be about half as wide as the ones in your example image. Is that adequate? Make sure you consider AJ Henderson's caution about motion blur: you will have to make sure the camera is steady in order to not have blurred details at that level of zoom.

I suggest to rent a P900 and test it, before you commit to buying. For around $60, you can rent one from places such as LensRentals.com or BorrowLenses.com for a week.

Originally by user11924. Source · Licensed CC BY-SA 4.0

user11924

9y ago

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AI Answer

Generated from our catalog & community — verify before relying on it.

Don’t judge by “24x” or “60x” zoom. That number only tells you the zoom range, not how much reach the camera actually has. What matters is the actual focal length, sensor size, and image resolution.

A practical way to estimate it is by scaling from a test shot: if a photo taken at 12 ft with a 300mm-equivalent focal length gives enough detail, then you’d need about 600mm at 24 ft, 1200mm at 48 ft, and 2400mm at 96 ft for the same subject size in the frame. That means the worst-case distance may require extremely long telephoto reach.

So yes, a superzoom/bridge camera may be enough only if its maximum equivalent focal length is long enough for your required detail. The “x” rating alone is meaningless.

Best approach: use any camera you already have, take test photos at measured distances, note the focal length used, and scale from there. This will tell you whether you only need to confirm the cable is intact or whether reading the tag is realistic. A polarizer may help reduce glare in daylight, but the key issue is focal length and resolution, not zoom ratio.

UniqueBot

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9y ago

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