Playing around with the incomparable lens calculator in Metric Mode I happened upon this equality:
When using a 1/3" sensor*, if the distance to subject (measured in meters) = focal length (measured in millimeters) then the HFOV is ~5m (4.8m to be exact). Could this simple equality be turned into a rough estimation rule of thumb? Your thoughts and please feel to be as critical as I deserve. Here goes:
A Ford Ranger (Regular Cab),
is also about ~5m (4.8m) long. Therefore the image frame would be completely filled (across) by the truck's profile if one were using a:
- 10mm lens when the truck is 10m away
- 25mm lens when the truck is 25m away
- 80mm lens when the truck is 80m away
Easy to remember. Said differently, when using a 1/3" sensor, if the number of meters to the subject is the same as the number of millimeters in the lens length, then the HFOV at that distance is always the size of a Ranger.
Ok, big deal you say, what good is that when my focal length and distance to subject are not equal, like 99% of the time? Because you can still get a rough idea of the area covered by the lens/camera very rapidly, and for multiple data points. It's also a good sanity check that you didn't make an entry error using a calculator.
So let's say you want to visualize the near coverage of a 15mm lens out into a parking lot**, one could imagine a Ranger traversely parked 15m from the lens. Make a triangle, starting at the lens, with two lines diverging to each bumper of the truck and then continuing to the horizon. Everything inside the triangle would be covered by the lens/camera.
Why a Ranger anyway, why not a Bronco? Because a Ranger is the same horizontal size in meters, 4.8, as a 1/3" sensor is in mm.
*Other formats have slightly different constants. For instance a 1/2.7" sensor would be 5.37M.
**I'm sure a large number of members can visualize camera/lens coverage areas precisely without the use of such devices, I'm not one of them, yet.