How far away is that plane?

[imamartian]

If i can see a plane in the distance (i presume cruising at 30,000 ft), how far in miles across the ground would it be from me?

I started trying to work it out, but not sure whether to use a sine or cosine (hmmm!).

What i want to understand is, as i live roughly in the middle of England, can i see all the air traffic up and down the country from the East to the West coast.

I know, i know, but the Nafta's were pretty boring and there's not much else on !

 

[^woody^]

You'll need at least one distance and/or the angle of view to work it out.

And don't forget to factor in the earths curve

 

[tim west]

It is a moving plane so how will you tell the distance

 

[Diyisfun]

Have a guess 30,000 feet convert to miles.
Then time how long the plane takes to get over your head & guess the speed, then you can calculate how far away IT WAS

 

[Nige F]

What i want to understand is, as i live roughly in the middle of England, can i see all the air traffic up and down the country from the East to the West coast.

!

have you seen any towing the remains of a conveyor belt....and the pilot gives you " The Finger"

 

[megawatt]

This one is easy to work out as long as you remember the rule ...

The angle of the dangle = twice the height of the flight divided by the elasticity of atomicity.

Don't you guys know ANYTHING ... Geeeez!

MW

 

[noseall]

^woody^ hit the nail on the head when he mentioned the earths curvature.
plus, atmospheric conditions normally prevent long distance viewing, even on a clear day. looking straight up is ok, but looking across diagonally is restrictive.

for example, a large tall ship will only be visible by the very top of its mast after only after 10 miles or so.

 

[Thermo]

could you not work it out roughly by timing it once it has passed overhead. You can probably make an assumtion about speed (ish sort of if your sad enough to know what plane it is, but the your sad enough to be trying to work this out)

 

[imamartian]

and i thought i was gonna get my trig done for me.... no such luck on this site

i thought -

tan 30deg = opposite / adjacent

= height / distance away

= 30,000ft / distance

not sure how to solve that, but am i way off the mark? And what's the curve of the Earth got to do with it?

 

[Thermo]

no curve, the earths flat everyone knows that

 

[noseall]

for triginometry to work, you need at least two points of reference, whether they be an angle or distance.

if the plane is 30,000 (5.5 miles or so) feet up vertically and is 52,800 feet away horizontally (10 miles or so) then it is 60200 feet away diagonally (11.4 miles or so)

no need for trig, just pythag.

 

[imamartian]

noseall, i am trying to work out the horizontal distance based on the height(30,000ft) and the viewing angle.... 20-30 degrees.

Your example didn't compute either.... a triangle with sides 5.5miles and 10miles would have a hypotenuse of 11.4miles (don't mean to be picky )

 

[noseall]

i've edited due to stella.

 

[imamartian]

Yeah i know stella, she not good at maths !!

 

[^woody^]

Ok this does not allow for the earths curve, and assumes a right-angled triangle

You will need to know 3 things for solving trigonometry - either 3 sides, 2 angles and a side, or two sides and an angle

In this case we know the height, but need to know the angle from the observer (A) to the plane, and distance to the plane (hypotenuse) (D)

Ground distance = D cosA