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Originally Posted by said
'Off on a tangent' you mean!
There's a more obvious reply than that .
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Care to have another go? .
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Originally Posted by said
You asked? This is not going to be easy - I have not figured out the new software I am using to accept drawings - so, I will try explaining. I don't know how much you know, so I will cover everything regarding Pythagoras and the Right Angled Triangle.
If you have a line of say - 2 (cms, metres, miles, feet whatever units you are working with) then square it. 2 squared = 4, what you have done by squaring a line is that you have made a perfect square shape with every side the same size, 2. We will use metres for now. So every side of that square is two metres long. The area of that square is Length x Width = 2m x 2m = 4 square metres.
Whatever length you choose to use, by squaring it, y0u will have the same result. A line 3m long, when squared will have a square with sides of 3m long each side, and 3 squared is 9 square metres - which is also the area of the square.
If we look at a square, and if we cut it in half diagonally, we will have two right angled triangles. A right angled triangle has one 90 degree angle between the vertical and the horizontal pieces. The longest side of that triangle is called the hypoteneuse and this long side is ALWAYS opposite the 90 degree join on the triangle.
Pythagoras' Theorem is - by taking the long side of the triangle and forming that into a perfect square. Then take the smallest side of the triangle and make that into a perfect square, and also do the same for the other side of the triangle. You will find that by measuring each square - that the area of the two smaller squares added together is EXACTLY the same area of the larger square. You don't have to do anything with that - but just bear it in mind.
On a piece of paper, draw the sort of triangle you would get if you cut a square shape horizontally in half. A Right Angled triangle. Label the long side of this triangle as 'A', and the medium side as 'B' and the smallest side as 'C'. Now going by Pythagoras, where the square on the smaller sides of the triangle is equal to the square on the larger side, we now have side B squared + Side C squared = Side A squared.
To show we are squaring a number, use a small '2' near the top right of each letter: B^2 + C^2 = A^2.
Right you are standing on the beach at sea level. Below your feet, many kilometres down - is the centre of the Earth. For these calculations we will assume that the earth forms a true circle. The distance across the centre of any circle is its DIAMETER. The distance from the centre to any point on the edge of the circle is its RADIUS. You are standing on the beach (#Anywhere on the surface of Earth is on the edge of its circle) - the distance to the centre of the Earth from where you stand is its RADIUS. The Radius of the Earth is found on the internet and given as 6371km. There are 1000 metres in a kilometre - so in metres the Radius of the Earth would be 6371000 metres.
You can imagine a line from your feet to the centre of the Earth, and another line of your sight towards the horizon, and a third line from the object in your sight back to the centre of the Earth. These lines form a right angled triangle.
You are standing on the hypoteneuse of that triangle, because it is opposite the right angle formed from your sight line to the object of sight to the centre of Earth and it is also the longest side because the length of this line is the radius of Earth added to your height, or the height at which you choose to be. The lien of your sight to the horizon is distance 'd' and the other line to the centre of the Earth, its Radius we will call 'R'. So by Pythagoras, we have that d^2 + R^2 = (R + h)^2
If you have followed that very long winded explanation you have done well - we can now find the length of the side of this triangle that corresponds to you sight line, which is line 'd'.
(R + h)^2 can be expanded to the form R^2 + 2 Rh + h^2. This gives the formula above as
R^2 + 2Rh + h^2 = d^2 + R^2 The R^2 on each side cancel out.
So we have:
2 Rh + h^2 = d^2 In comparison to the other values in the calculation, h^2 is so very small, it is neglible, so we will drop that term.
We now have:
d^2 = 2Rh^2
and d = sq.rt. 2Rh Where 'h' is your height, either on the beach or if you choose at an elevation, and R is the radius of the Earth.
If you do that - I can show you how to find the depth at which you can see part of your object. Though you probably know now.
My brain hurts.
Thank you very much for that, Said. It must have taken you a while to compose. I have printed it off on to paper so I can study it properly as my eyes being to strain when I'm concentrating on a screen .
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Originally Posted by TownieChap
My brain hurts.
Thank you very much for that, Said. It must have taken you a while to compose. I have printed it off on to paper so I can study it properly as my eyes being to strain when I'm concentrating on a screen .
That is OK - hopefully you will be able to follow the steps, if you do manage to put your measurements into the formula, you should get a good result - any problems, send me a PM.
As you have realised, the higher you stand the greater distance you can see to the horizon. Climate conditions also affect the distance that the horizon appears to be at.
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Originally Posted by Desert Region
Care to have another go? .
I know exactly what a tangent is, if you mean mathematically - but there was no need to use that in this formula. There are several formulas that can be used, including those that involve refraction with temperature gradients etc., but for general purposes this formula suits.
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Originally Posted by said
I know exactly what a tangent is, if you mean mathematically - but there was no need to use that in this formula. There are several formulas that can be used, including those that involve refraction with temperature gradients etc., but for general purposes this formula suits.
Actually, your line of sight 'd' in this formula IS a tangent to the earth.
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Originally Posted by Snig's foot
Actually, your line of sight 'd' in this formula IS a tangent to the earth.
Obviously you have never done teaching. Working with the tangent from line of sight to the horizon, involves either working with the secant/tangent formula and vectors or working with algebra and a knowledge of the various forms of differentiation. Would you find that easier?
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Originally Posted by said
Obviously you have never done teaching. Working with the tangent from line of sight to the horizon, involves either working with the secant/tangent formula and vectors or working with algebra and a knowledge of the various forms of differentiation. Would you find that easier?
yes please.
"May have been the losing side. Still not convinced it was the wrong one." - Firefly (TV Series)
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Originally Posted by AdmiralAckbar
yes please.
It's just marvellous, isn't it? .
I'm also hoping to see instructions provided on how to estimate the depth of a very deep hole.
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Originally Posted by Desert Region
It's just marvellous, isn't it? .
I'm also hoping to see instructions provided on how to estimate the depth of a very deep hole.
This guys got it covered
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Originally Posted by Desert Region
It's just marvellous, isn't it? .
I'm also hoping to see instructions provided on how to estimate the depth of a very deep hole.
Providing you have the volume of earth taken from the hole - it is easy.
From Pi x radius of the hole squared x height, we get the height as volume of earth removed in cms cubed, divided by Pi x radius squared.
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Originally Posted by Desert Region
It's just marvellous, isn't it? .
I'm also hoping to see instructions provided on how to estimate the depth of a very deep hole.
I'm still busy trying to "draw the sort of triangle you would get if you cut a square shape horizontally in half. A Right Angled triangle ". So far I've only been able to draw a rectangle. What am I doing wrong? Does it involve the "hypoteneuse" (sic) ?
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Originally Posted by TownieChap
My brain hurts.
Thank you very much for that, Said. It must have taken you a while to compose. I have printed it off on to paper so I can study it properly as my eyes being to strain when I'm concentrating on a screen .
I have just had it pointed out that I have stated "cutting a square in half horizontally" When this should have obviously read "cutting a square diagonally" To give a triangular shape. Some people get really excited in being able to point out the obvious.
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Originally Posted by seivad
I'm still busy trying to "draw the sort of triangle you would get if you cut a square shape horizontally in half. A Right Angled triangle ". So far I've only been able to draw a rectangle. What am I doing wrong? Does it involve the "hypoteneuse" (sic) ?
Did it not occur to you that this may have been an error? It is obvious that you do not get a triangular shape if you halve a square horizontally. I think most people would have realisedthat it should have read diagonally(erata corrige)
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Originally Posted by said
Did it not occur to you that this may have been an error? It is obvious that you do not get a triangular shape if you halve a square horizontally. I think most people would have realisedthat it should have read diagonally(erata corrige)
Of course I realised that it was an error. If you read over your post before submitting it, you would have spotted it too!
You'd better add "hypoteneuse" to your errata... and "erata" too!
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