Determining Distance to Object, Retina Image
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?
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.
John C. Strong
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
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! =)
Click here to return to the Physics Archives
Update: June 2012