Imagine a car or truck speeding away in front of you on the road. Here is a rough method to estimate its distance and speed using your outstretched hand (try this as a passenger, not a driver!).
At any given time its distance away is approximately
We use the figure below to estimate angles based on our outstretched hand at arm’s length:
Hence a vehicle that is 2m wide subtending an angle of 5 degrees (3 fingers at arm’s length) is about 23m away. At 1 degree (1 little finger at arm’s length) it is about 115m away.
Then measure the time T (in seconds) it takes to go from 3 fingers width (5 degrees) to 1 little finger width (1 degree) at arm’s length. Its relative speed in km/h is approximately given by
For example, if a 2m-wide car takes 33 seconds to go from 3 fingers to 1 little finger width at arm’s length (115-23=92 m in distance), it is going 10km/h faster than I am.
To justify these formulas, the figure below shows segment BC of width w at distance d from A, subtending an angle of 
Formula (1) then follows from 
Formula (2) is true since we are covering a distance 
The table below shows some sample values for a car with width 2m and truck with width 3m.
| Angle subtended by object (degrees) | Width of vehicle (m) | Distance of object (m) | Approximate distance using (1) |
| 20 | 2 | 5.7 | 5.7 |
| 10 | 2 | 11 | 11 |
| 5 | 2 | 23 | 23 |
| 1 | 2 | 115 | 114 |
| 0.02 | 2 | 5730 | 5700 |
| 20 | 3 | 8.5 | 8.6 |
| 10 | 3 | 17 | 17 |
| 5 | 3 | 34 | 34 |
| 1 | 3 | 172 | 171 |
| 0.02 | 3 | 8600 | 8550 |
The smallest angle of 0.02° (~1 minute) approximately corresponds to the smallest object we can resolve with the naked eye.
The next table shows estimates of receding speeds based on going from 5° to 1° of angular diameter.
| Width of vehicle (m) | Time (s) to go from 5° to 1° | Estimated speed (km/h) |
| 2 | 3 | 110 |
| 2 | 5 | 66 |
| 2 | 10 | 33 |
| 2 | 33 | 10 |
| 3 | 5 | 99 |
| 3 | 10 | 50 |
| 3 | 33 | 15 |
Reference:
. Assume the observer is in the northern hemisphere and the point 
represents the north celestial pole about which the stars appear to rotate anti-clockwise as the earth rotates on its axis (in the southern hemisphere stars rotate clockwise about the south celestial pole). The red ellipse illustrates the circular path of a star over a 24-hour period – point 
is its highest location while point 
is where it will be after a quarter of a day. Also shown in the diagram are axes 
(pointing east), 
(pointing north), 
(pointing directly overhead) and corresponding unit basis vectors 
, 
and 
.
– the time (as a proportion of a day) after the star has reached its highest point
– the angle of 
– the latitude of the observer (-90° to 90°)
. If 
, we would have a northern pole star and 
. If 
we would have a star on the celestial equator and 
. More generally,
, hence we have
from the fixed line 
which is the projection of 
and 
. For any point 
on the path we can write![\displaystyle \vec{A^{'}P} = \cos 2\pi t\ \vec{A^{'}B} + \sin 2 \pi t\ \vec{A^{'}C}, \quad t \in [0, 1].\quad\quad ...(6)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cvec%7BA%5E%7B%27%7DP%7D+%3D+%5Ccos++2%5Cpi+t%5C+%5Cvec%7BA%5E%7B%27%7DB%7D+%2B+%5Csin++2+%5Cpi+t%5C+%5Cvec%7BA%5E%7B%27%7DC%7D%2C+%5Cquad+t+%5Cin+%5B0%2C+1%5D.%5Cquad%5Cquad+...%286%29&bg=ffffff&fg=000000&s=0&c=20201002)
, 
, 
, 
, 
has the same length as 
and is pointing directly west (i.e. perpendicular to both 
and 
), we have from (5):
:
equal to the 
, the fixed location of the pole star.
. This traces a circle with diameter joining 
and 
(two antipodal points), hence it is a great circle on the celestial sphere.
(observer on the equator) we obtain 
, hence the 
(observer at the north pole) we obtain 
, hence the star traces out a circle with constant angle of elevation.
(assuming a circular orbit with radius 
starting at 
at 
). The parameter 
here represents the fraction of year after the winter solstice. Earth’s axis of rotation points in the direction 
, being tilted away from the sun in the north when 
.
. Then the angle 
between the sun’s position and the earth’s axis satisfies
. If 
, we have the equinoxes and 
then 
.
and hence
. For example taking the sun on the summer solstice at Melbourne, Australia we have 
and so 
. This corresponds to a bearing of 
, around 20 degrees south of east. The answer given in 
, slightly more perhaps because it takes into account refraction and the fact that the sun is not pointlike.
are to be determined. Multiplying both sides by 
for some 
,
in both sides gives
![\displaystyle \begin{aligned}n=1&:& \frac{1}{x(x+1)} &= \frac{1}{x} - \frac{1}{x+1}\\n = 2&:& \frac{1}{x(x+1)(x+2)} &= \frac{1}{2} \left[ \frac{1}{x} - \frac{2}{x+1} + \frac{1}{x+2} \right]\\n=3&:& \frac{1}{x(x+1)(x+2)(x+3)} &= \frac{1}{6} \left[ \frac{1}{x} - \frac{3}{x+1} + \frac{3}{x+2} - \frac{1}{x+3}\right] \end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7Dn%3D1%26%3A%26+%5Cfrac%7B1%7D%7Bx%28x%2B1%29%7D+%26%3D+%5Cfrac%7B1%7D%7Bx%7D+-+%5Cfrac%7B1%7D%7Bx%2B1%7D%5C%5Cn+%3D+2%26%3A%26+%5Cfrac%7B1%7D%7Bx%28x%2B1%29%28x%2B2%29%7D+%26%3D+%5Cfrac%7B1%7D%7B2%7D+%5Cleft%5B++%5Cfrac%7B1%7D%7Bx%7D+-+%5Cfrac%7B2%7D%7Bx%2B1%7D+%2B+%5Cfrac%7B1%7D%7Bx%2B2%7D+%5Cright%5D%5C%5Cn%3D3%26%3A%26+%5Cfrac%7B1%7D%7Bx%28x%2B1%29%28x%2B2%29%28x%2B3%29%7D+%26%3D+%5Cfrac%7B1%7D%7B6%7D+%5Cleft%5B+%5Cfrac%7B1%7D%7Bx%7D+-+%5Cfrac%7B3%7D%7Bx%2B1%7D+%2B+%5Cfrac%7B3%7D%7Bx%2B2%7D+-+%5Cfrac%7B1%7D%7Bx%2B3%7D%5Cright%5D+%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0&c=20201002)
is drawn parallel to 
), thus dissecting the triangle into two. Each of the smaller triangles then has an inscribed parallelogram where two of the vertices are on a side of each triangle. Then by drawing lines parallel to sides if required, we create two sub-problems each having four vertices of the parallelogram on the sides of the triangle. 
formed by lines parallel to the sides.
as shown in the figure below). In this figure, if we let 
and the height of ABC from 
, then by the similarity of triangles 
and 
, 
and 
has height 
. Then the area of the parallelogram 
is 
which is 
times the area of 
. This quantity has maximum value 1/2 when 
so we conclude that the parallelogram does not exceed half the triangle’s area.
are the sides of the triangle with 
the second longest side length. Then let the segment 
joining the midpoints of 
and 
. It remains to be shown that there exist parallel segments of this same length from 
to 
. The longest possible such segment has length 
and the shortest has length 
, half the length of the altitude of 
. This follows from 
of 
inscribed in 
is on 
is on 
. We claim that the largest such square is either 
(two vertices on 
inscribed in 
, 
and 
. To show these circles intersect at a single point, we can prove that if the circumcircles of triangles 
are cyclic by the following equality:
, 
being cyclic. 
and 
. Hence 
respectively on the sides 
of the triangle. Consider the locus of the points of the square as 
also traces a line segment, between the points 
and 
so as to be contained within the triangle. 
is maximised. We have seen that the point 
or 
. We therefore conclude that the largest square inside a triangle will have two points on a side. 
of a square on the side of length 
with altitude 
is the same as if 
, where 
is right-angled. In this case the square’s side length is 
.
is obtuse
.
and hypotenuse 
, we wish to compare the quantities 
and 
, the two possible sums of the base and height of the triangle. We always have 
because the diameter of the incircle of the triangle is shorter than the altitude from the hypotenuse (i.e. the incircle is inside the triangle). We conclude that the largest square in a right-angled triangle is constructed on its two legs rather than its hypotenuse.
(midpoint of the midpoints of AB and BC)
(where 
)
and 
): 
and 
: 
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