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Name: Richard
Status: other
Grade: other
Location: RI
Country: USA
Date: Fall 2011

What is the formula for determining the distance away an object is, given the apparent size versus the actual size? For example, an airplane flying overhead. I live near an airport and curiosity got me thinking about this. It might be one inch to my retina versus say, a thousand feet long actually. What is it's distance away?

Hi Richard,

You can first estimate the angular diameter of the object in your view. That is the length or diameter of the object in terms of angular units (e.g., degrees). For example, both the moon and the sun are about 0.5 degrees in diameter from our vantage point on Earth. The distance from horizon to horizon would be 180 degrees. Estimating the angular diameter may not be very easy and can be prone to error. Since the moon is very familiar, you can estimate the object in terms of how many moons in length/diameter it is, then convert to degrees.

Now if you know the actual length or diameter of the object you are looking at, you can use simple trigonometry to figure out the distance. The tangent of half the angular distance you estimated above (let us call that "a") would be equal to half the actual diameter or length (let us call that "d") divided by the actual distance (let us call that "D").

So, if we draw out the triangles we can see that tan(a/2) = (d/2) / D

or, D = (d/2) / tan(a/2)

For example, we know the moon is about 3476 km. So then D = 1738 km / tan(0.25 deg) = 398,000 km. And that turns out pretty close to the known average distance to the moon which is 384,000 km.

You can use this same procedure another way. Say you hold your left arm straight out in front of you and stick out your index finger to the right so you can see it clearly. You measure the distance from your eye to your finger and it is 45 cm. You measure the thickness of your finger and it is 1.5 cm. This means there is a factor of 30 when you use your finger as a measuring guide. So if you see an airplane fly directly overhead, and you use your measuring finger to see that your finger just barely covers the wing, and you know the wing is 33 foot span, then you can quickly estimate that the airplane is about 1000 feet overhead. If the plane is two fingers in span, then its 500 feet above you. If half your finger is enough to cover the plane, then its 2000 feet above you.

Regards, John C. Strong

Hello Richard,

Good observation.

It is a matter of the ratio of triangles where:

The ratio of the perceived dimensions = The ratio of the real dimensions.

First you calculate the perceived dimensions, and since you know one "real" dimension ( the length of the plane), you can then solve for the last dimension - the distance.

Plot on a sheet of paper the flight path of the plane overhead, and note each time the plane travels its length as perceived by your eye, until you record, say about 10 of them.

Then, draw a line from the point of origin, where the plane started overhead, to this 10th point you plotted and label this line X. Measure and record the length of X. If to your eye, the plane is 1" long, then your line X should be approximately 12".

Then on that same sheet of paper, draw a vertical line from the point where the plane started overhead, down to your home, and label that distance Y.

Next draw a line from the 10th point the plane traveled to your home and label that line H.

Now measure the angle between the vertical line Y and the line H. You can call this angle Theta. The reason for plotting 10 points is to get a good angle measurement.,

Also from Trig, the Tangent of an Angle = distance opposite the angle / distance adjacent to the angle. The distance opposite the angle is the distance the plane traveled, and the distance adjacent is the distance over your head. So in your picture, tan(Theta) = X/Y.

If you perceived the plane length to be 1"' then the measured distance traveled, X, on your paper, should be 12". From there, then the distance overhead can be calculated by rearranging the equation above to solve for Y:

Y = X / tan(Theta)

Now the matter of ratios; but before that, we have to divide X by 10. Remember we plotted 10 plane lengths to get a good measurement of angle Theta.

Equate Z=X/10, in your case Z=1.2"

Now we have dimensions for two triangles. One the perceived triangle and the other the "real" triangle.

The ratio of the perceived dimensions (Z/Y) is equivalent to the ratio of the "real" dimensions (Plane's length / Distance overhead).

Rearranging, we get: Distance Overhead = Plane's Length / (Z/Y)

With that, we get Distance = 1000 feet / (1.2/20) is approximately 16,000 Feet.

Sounds about right given the floor (minimum height) restriction for a residental area is typically 10,000 to 15,000 feet. You can confirm by asking any pilot or look at flight maps, or call the airport! =)

Stay Curious! -Alex Viray

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