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Screwballs
14th Jan 2009, 13:22
This is an obscure question so it's only out of curiousity I ask:

How can you work out how much distance you travel towards a position directly below you as you move around a circle?

So for example, if you move from A to B and B is one quarter of the way around the earth, for example, you travel 5400 nm great circle and you move vertically downwards by 3437.74 nm (or one radius of the earth). In other words you could travel directly down to the center of the earth at point A and then travel forward towards B and you'd end up at B (vertical) having travelled 3437.74 nm "down".

Does the question make any sense?

Thanks for any replies,

Screwballs (who needs to get out more)

Dit
14th Jan 2009, 13:38
If you're asking what I think you are, whcih is basically how do you work out the circumference of a circle, it'll be Pi multiplied by the twice the radius of the circle. In your exaple your asking for quater of the circumference so it'd be:

(Pi x 2r)/4
(3.14159 x 2 x 3437.74)/4 = 5399.989

Or you could be asking a different question, in which case I don't have a clue what you're on about.

pilotmike
14th Jan 2009, 15:05
Does the question make any sense?
No. You have a most baffling and bizarre way of thinking... I believe this one will travel 'great circle' to Spotter's Balcony or Jet Blast at aproximately the speed of light.

However, in an effort to answer the question which I believe you were trying to ask, your 'height loss' or 'vertical distance' is given by this cunning formula, which I have devised especially for you on your journey to the centre of the Earth:

'height loss' = 3438NM x (1 - COSINE ('distance travelled' / 60)), which works for any distance travelled, whether part way round the Earth, completely round it, and even going around it as many times as you care to go round in circles.

The above formula assumed Degrees in the COSINE function. If you need to use Radians on your unbelieveably bizarre journey, then the following formula may be used:

'height loss' = 3438NM x (1 - COSINE ('distance travelled' / 3438))

Now, by return, would you kindly answer for me, what relevance this is to anyone, for any purpose whatsoever, please??!!!

TheGorrilla
14th Jan 2009, 16:14
Is screwballs planning on going into space?

LH2
14th Jan 2009, 16:44
How can you work out how much distance you travel towards a position directly below you as you move around a circle?

You need to define a suitable reference system for your question, such as a three-dimensional earth-fixed cartesian system with its origin at your point of departure, and its Z axis aligned with the local vertical at same. Then your "distance [...] towards a position directly below you" is simply the value of Z at your destination, and the "distance you travel [along Z] towards a position directly below you" can be found by integrating |Z| along your route.

Does this answer your question? Sorry, but the phrasing in your original post was a bit vague so I'm just having a best guess at what you were trying to ask.

Screwballs
15th Jan 2009, 09:29
This is an obscure question so it's only out of curiousity I ask:

I would have presumed that the first line in my thread would have covered the whys and wherefores however...

Thanks pilotmike for the formula, it was just idle curiosity and I couldn't work one out myself!

Screwballs

Rainboe
15th Jan 2009, 14:25
No it doesn't actually. It's such an obscure, nonsensical question with no logic I read it several times and still could not understand it! I am surprised anybody answered it! So, no, your first line does not actually provide an exoneration! I think my first wife has similar confused thought processes.

Chesty Morgan
15th Jan 2009, 18:21
How can you work out height loss relative to your starting position?

Answer - Subtract your finishing position from your starting position.

How can you work out how much distance you travel towards a position directly below you as you move around a circle?

Answer - Spherical trigonometry - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Spherical_trigonometry)

pilotmike
15th Jan 2009, 18:24
No, Screwballs, Rainboe is completely right. It really was a truly hairbrained question, and worded in such poor layman's terms, that I found it hard to believe that you are indeed a pilot.

Just because Rainboe points this out shouldn't cause you to get your knickers in a twist with him!

Screwballs
15th Jan 2009, 18:33
Ah come now, not got the knickers in a twist, just trying to wind him up as he does to everyone else.

I thought there was no such thing as a stupid question but I guess I was wrong. Better stop asking anymore questions then!

Thanks for the answer all the same.

Screwballs

pilotmike
15th Jan 2009, 18:42
No sweat, Screwballs! I never meant to have a pop at you, I just had to slip 'knickers' into a post to win a £20 bet - honestly!

For my next £20, - 'dehumidifier'.

There, I've now earned 40 quid in just 20 minutes, and I haven't even gone out to work yet, I report in 3 hours time.

Hot 'n' High
15th Jan 2009, 19:22
LOL Screwballs – interesting question - with some merit (and controversy) too! My 0.5 penny’s worth! Not quite sure (as a simple bloke) where your “..around the earth” dit comes from – but, as a some-time simple glidist, I often have to work a similar question out for real - when looking for a rescue-thermal to save my bacon. My question in a dying thermal is “How long have I got before I have to commit to a ‘Downwind to a field landing’ decision?”

The simple answer I apply is ((Current height agl (estimated from topography) – 800ft agl)/ROD (ft/min)) = time left before I have to give it up and head for the sheep/cow/barley field! I know (roughly) how many RPM I am doing so I could work out how many orbits I could make. So, in theory, you could multiply TAS (GS should even out as constantly flying “into” and “out of” wind) by the time to “downwind decision point” or “crash site” to give a distance travelled to either event!

But, practically, I just work on a closest cut-off altitude to a downwind leg. When I reach that – "Hello sheep/cows/barley"!!!!! As for distance travelled? Mmmm, don’t care really as a pilot – but, using a formula for a conical spiral (or my calculation above), I guess you could work it out if aiming at a particular sheep within said field! However, in practical terms – neither I nor the sheep would be worried – provided (from both sides) I missed the sheep on flare/roll-out!!!!

Hope this is of mild curious interest! If not – my apologies! Ho hum!!!

H 'n' H