Calculating distance to the horizon
Digital Online Philomath
sqrt(F) * 1.225 = M, where F is the height in feet of the ground and M is
the distance in miles?
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Digital Online Philomath (pub...@jcrd.org)
Richard Bullock
"Digital Online Philomath" <pub...@jcrd.org> wrote in message
news:Xns92215F1169ADdi...@207.218.245.68...
> Is this formula for calculating the distance to the horizon correct:
> sqrt(F) * 1.225 = M, where F is the height in feet of the ground and M is
> the distance in miles?
Assuming a perfectly spherical Earth (not true) of radius r metres and an
observer whose eye is h metres off the ground, then the angle at the centre
of the Earth between where that observer would measure the horizon and the
observer, x can be summed up in the following
cos x = r / (r + h)
x = arccos [r/(r+h)] (where arccos function is in radians)
The curved distance, D from the point on the Earth directly below the
observer to the observer's horizon would then be given by
D = rx
D = r arccos [r/(r + h)]
The straight line distance, d, from the observer's eyes to the observer's
horizon is also given by (by Pythagoras' Theorem)
d = sqrt(2rh + h^2)
For small values of h, this simplifies to d = sqrt(2rh)
For d in miles and h and r in feet, this boils down to
d = 1.2252 * sqrt(h)
e.g. For an observer standing on the Earth (r = 6378km) with eyes at 1.75
metres (5' 9")
Curved distance, D = 4.725 km (2.936 miles)
The other formula gives the same distance for straight line distance.
Your formula gives 2.935 miles for the distance to the horizon - very good
approximation.
Try another example
An aircraft is flying at 10,000metres (32,800 ft)
My formula, D = 356.92 km (221.78 miles); d = 357.20km (222.01 miles)
Your formula gives d = 221.89 miles - still a good approximation - only 200
yards out in over 200 miles.
A further example.
An orbiting spacecraft at 800km (500 miles) above the surface of the Earth.
My formula, D = 3,039.9 km (1,888.9 miles); d = 3,293.1 km (2,046.3 miles)
Your formula gives d = 1,984.6 miles - approximation begins to break down
Taking it to the extreme.
From the Moon, how far is Earth's horizon.
My formula, D = 9,914km (6,160 miles) ~ 1/4 of Earth's circumference.
d = 390,326 km (242,537 miles) - not surprising - this is just over the
distance from Earth to the Moon.
Your formula gives d = 43,480 miles - very large error!
To sum up, yes, your formula is good enough for most heights that you choose
to calculate, but the approximation gets worse as you go higher up.
Ric
Digital Online Philomath
> To sum up, yes, your formula is good enough for most heights that you
> choose to calculate, but the approximation gets worse as you go higher
> up.
Hey, thanks for all of that! I'll start using YOUR formula instead.
Richard Bullock
> // Richard Bullock / Sun, 02 Jun 2002 15:05:44 GMT //
>
> > To sum up, yes, your formula is good enough for most heights that you
> > choose to calculate, but the approximation gets worse as you go higher
> > up.
>
> Hey, thanks for all of that! I'll start using YOUR formula instead.
> --
Pythagoras's theorem. For heights of hills/mountains, your formula does have
a good approximation. With my formula, you can transpose it to other
spherical objects of different sizes. e.g. working out the distance to the
horizon on the Moon that the Apollo astronauts saw.
You only have to remember that the distances all have to be in the same
units. So if you use metres for h, then you must use metres for r and the
resulting distance will be in metres.
Ric
RA
pub...@jcrd.org says...
> Is this formula for calculating the distance to the horizon correct:
> sqrt(F) * 1.225 = M, where F is the height in feet of the ground and M is
> the distance in miles?
>
I say yes.
--
Night Boar
(Portugal)