Black Combe and Snaefell | Why can we see one and not the other?

[TownieChap]

Standing on Southport Beach on a clear day, as I'm sure we've all done, you can see the peaks of the lake district providing an epic backdrop to Blackpool Tower as you look over to the fylde coast. The most prominent feature of the lake district seen from Southport Beach is Black Combe (It's the very big hill, not quite a mountain and technically called a Marilyn, that is seen to the West of Blackpool as we look at it). Black Combe stands at a smidge over 2,000 feet and as the crow flies it is located 43.93 miles from the Trans-pennine trail start point which is located on the slip road of our beach.


Now then. On that same clear day looking out to the Irish Sea in the general direction of the Isle of Man, can we see the Isle of Man? Nope. The highest point of the Isle of Man is Snaefell standing at 2,037 feet and also considered a Marilyn and is 72.49 miles from the same point on Southport Beach. Snaefell is 28.56 miles further away from us than Black Combe but stands at an equally considerable height.

I am hoping that somebody cleverer than me can explain how the curvature of the earth at 28.56 miles can effectively 'hide' something so large. If Black Combe where 28.58 miles further away from its location, surely we would still see some of it.


Attachment 189593

[TownieChap]

Sidenote: I have used strong binoculars on a very clear day and the Isle of Man can not be seen. The furthest landmark that can be seen is the Morecambe Bay Gas Platform.

[Hamble]

Is it to do with sea level?

Southport is 12m above.

IOM 620 m

[Knot wright]

this chap has done a nice little calculator

http://www.ringbell.co.uk/info/hdist.htm

if we set the observation height to 2000 feet , the horizon is some 55 miles away , sort of between the 2 limits of distance

so by massaging the figures

Black Combe , you can still see the top 700 feet of it .

Snaefell fails to peep over the horizon , it would need to be over 3500 feet tall

[TownieChap]

this chap has done a nice little calculator

http://www.ringbell.co.uk/info/hdist.htm

if we set the observation height to 2000 feet , the horizon is some 55 miles away , sort of between the 2 limits of distance

so by massaging the figures

Black Combe , you can still see the top 700 feet of it .

Snaefell fails to peep over the horizon , it would need to be over 3500 feet tall

Ah! Brilliant. I did wonder about the height when I thought about Snowdonia in North Wales as that is 51.76 miles away from the same point as above but stands at around 3,5000 feet at highest point.

It's amazing to me that such a relatively short distance of 28.56 miles can eliminate the view of such a large 'Marilyn'. I will have to brush up on my understanding of distance and perception. Thank you very much for that, Knot Wright. Good work!

[said]

Ah! Brilliant. I did wonder about the height when I thought about Snowdonia in North Wales as that is 51.76 miles away from the same point as above but stands at around 3,5000 feet at highest point.

It's amazing to me that such a relatively short distance of 28.56 miles can eliminate the view of such a large 'Marilyn'. I will have to brush up on my understanding of distance and perception. Thank you very much for that, Knot Wright. Good work!

Or, alternatively - you could find a viewing spot at a greater height. Based on the calculations given - you would need to be about 460/500 feet higher up - I don't think Parbold Hill is quite high enough, but Winter Hill may be.

[chrismatt.]

From the top of Winter Hill you can see the Isle of Man: link below

https://www.flickr.com/photos/59251448@N04/25708799453

[TownieChap]

From the top of Winter Hill you can see the Isle of Man: link below

https://www.flickr.com/photos/59251448@N04/25708799453

That's a great photograph. A trip to Winter Hill has just found it's way on to my to-do list. Great stuff!

[said]

That's a great photograph. A trip to Winter Hill has just found it's way on to my to-do list. Great stuff!

You going to be doing this a lot? Are you OK with calculations to work out height and distances above and below horizon, Pythagoras? Or do you need example to refresh?

[TownieChap]

You going to be doing this a lot? Are you OK with calculations to work out height and distances above and below horizon, Pythagoras? Or do you need example to refresh?

A Pythagoras what? It sounds like something I tried to order in a Greek restaurant once after drinking one to many glasses of lovely wine!

I shouldn't imagine I'll be doing a lot of standing on high lands looking out onto the horizon in awe and wonder. That said, I will certainly take a trip to Winter Hill with my nosey-man's binoculars. I am a man of short lived varying hobbies and interest. Last month I took an overwhelming interest in the different size of railway gauges used all over the world, the month before that I had flirt with Aviation and how air traffic control systems work... so you see, varying interests! Feel free to pass on what you know about Pythagoras. I'm all eyes!

[Kritou]

ii’s not just the earth’s curvature that affects how far away we can see things but factors such as atmospheric pollution and moisture content (think desert mirages) are also contributors

A friend who lives near the seafront in a flat that is probably about 95ft AMSL - from there we frequently see Snaefell and with binoculars have even made out the 2500ft peak just south of Newcastle in Northern Ireland

[TownieChap]

ii’s not just the earth’s curvature that affects how far away we can see things but factors such as atmospheric pollution and moisture content (think desert mirages) are also contributors

A friend who lives near the seafront in a flat that is probably about 95ft AMSL - from there we frequently see Snaefell and with binoculars have even made out the 2500ft peak just south of Newcastle in Northern Ireland

That's brilliant, Kritou. I would never have thought that Northern Ireland would be visible at all, and yes, you are right. There are lot's of factors that can contribute to the view of an horizon. I am learning all about them as this thread goes on

[said]

A Pythagoras what? It sounds like something I tried to order in a Greek restaurant once after drinking one to many glasses of lovely wine!

I shouldn't imagine I'll be doing a lot of standing on high lands looking out onto the horizon in awe and wonder. That said, I will certainly take a trip to Winter Hill with my nosey-man's binoculars. I am a man of short lived varying hobbies and interest. Last month I took an overwhelming interest in the different size of railway gauges used all over the world, the month before that I had flirt with Aviation and how air traffic control systems work... so you see, varying interests! Feel free to pass on what you know about Pythagoras. I'm all eyes!

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.

[Desert Region]

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.

In brief: I see someone's gone on a tangent!

[said]

In brief: I see someone's gone on a tangent!

'Off on a tangent' you mean!