Finding the sizes of – and distance to – distant targets (or “Why Fernieboy100 is menatlly challenged”).

In this post, we’re going to discover just how easy it is to measure the distances and sizes of objects, without having to walk to them.

It’s an inverted triangulation method. Instead of triangulating your position, you can triangulate a chosen object.

YouTube user Fernieboy100, once asserted that we couldn’t measure the distances to objects using telescopes or binoculars, during one of his least mentally attuned attempts to debunk a video by CoolHardLogic.

CoolHardLogic‘s brilliant video response, on how we can measure the distance to the Sun, can be found here:

Indeed, Fernieboy100 seems to believe that we can only measure distances by getting a tape measure and walking towards our targets.

He’s obviously never met anyone who works with maps – and must be living under the delusion that maps are made by people walking all over the countryside with a massive ruler.

However, we’ll just need 2 angles and a baseline and we can calculate everything else.


We need a pair of binoculars/telescope, a pencil, a ruler, a protractor, a pocket calculator (or very good knowledge of maths), a sheet of paper and a magnetic compass. We also need to know our binoculars’/telescope’s field of view (normally written on the binoculars/scope), so we can measure the angular size of whatever object we like (a mountain, a tower, whatever).


First, we take a bearing to the object we want to measure the size of and distance to. Let’s say we pick a house.

Next, we’re going to draw a simple schematic map (schematic means it’s just a representation, we don’t need to draw the terrain).

We’re going to draw an arrow on our map to indicate which direction North is, and draw a long line in the direction of the object from a point that represents our starting point. We need to write on the bearing to the object as well.

Let’s say our house lies on a bearing of 120°.

Now we’re going to walk in any direction we want (that isn’t directly towards or away from the object), covering any distance we choose. A vaguely lateral path is best.

We only need to walk 500m or 1km at most, or if our compass has mil-radians marked on it (and we’re good enough at taking a bearing), we could even just walk 100m or less.

Before we set off, we’ll make a note of the bearing we’re walking in and draw a line on our schematic representing this. We’re going to scale the map as well. We can make 1cm=100m or something appropriate for the distance we’re going to walk.

In this case, we’re going to walk 500m on a heading of 210°.


I know my pace over 100m on a flat surface is 60 “right feet” (adjusting accordingly for terrain), so I can count out this distance, using my compass to stay on a straight bearing. The “right feet” unit means that I only count when I step with my right foot, which makes it easier to count over long distances. You can find out what your own measure is beforehand, by simply walking 100m and counting.


When we’ve reached our 2nd observation spot, we take a bearing on the object and draw a second line on our schematic map along that bearing from where we are now.

We can now know the angle described by the vectors emanating from our starting point to where we are now and the object, which is just the difference between the bearing from our first point and the bearing we’ve walked on.

We can also find the angles described by the vectors emanating from the object to our 2 observation points, as it’s just the difference between the 2 bearings.

Since the angles in a triangle must add up to 180°, we can also figure out the final angle at the point we’re now standing.


In this exercise, let’s imagine the bearing to the object from our 2nd point is 110°.

We know that if we’ve walked on a bearing of 210° from our first point to here, so the angle at our starting point is simply the bearing we’ve walked minus the bearing to the object from our starting point. 210-120 = 90°.

Since our original bearing was 120°, we can subtract our new bearing from this to find the angle at the object we’re measuring to be 10°.

180-90-10 = 80, so the angle at the point we’re standing at now is 80°. We only really need this final angle if we want to calculate the distance from our starting point to the object.


Because of the distances involved, we’re probably not going to have a piece of paper big enough to deal with the scales we’ve used, otherwise we could have just measured the distance with a ruler.

We’ll probably end up with something that looks more like this:


Which is why we’ll use trigonometry.


Now we have 3 angles and a length, we can calculate our distance to the object using the law of sines:

a/sin(A) = b/sin(B) = c/sin(C)

Rearranging this equation to find the length we want, we use:

a = (sin(A) x b)/sin(B)

Plugging in the numbers we’ve got, our distance is (sin(90) x 500)/sin(10), which gives us about 2879.4m, or 2.88km.

We can even find the distance from our starting point, if we wish. It’s (sin(80) x 500)/sin(10), which is about 2835.6m, or 2.84km.


Hey presto! We have the distance, thanks to a bit of maths and some basic tools.

Now we have the distance, we can find the size of the object.


Looking through our binoculars/telescope from the second point, we measure the angular size of the house to be about 11 arc minutes (if we’re good enough, we can even calculate the arc seconds for a better precision). Dividing our arc minutes by 60, we find that that’s 0.1833°.

The equation for finding the size of an object from its angular size is:

S is the size of the object.

a is the angular size of the object.

D is the distance to the object.

Plugging our numbers in, we find tan(0.1833/2) x 5857.8 comes to about 9m.


So, we’ve managed to measure the distance to, and size of, any distant object on earth using just binoculars/telescopes and some very basic trigonometry.


Now that we have the size of the object, if it is an object of roughly standard size, we can find other such objects and reverse the angular size equation to find our distances to them, using:

 D = S/(2 x tan(a/2))


Now we can easily measure the distance to anything and anywhere, using simple trigonometry and tools, to an acceptable level of accuracy.


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