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"Awwww the sea!! Happyface " Thank you, insightful as always... actually, I was relying on you to have the answer... bugger.. thats that plan well and truly scuppered.. | |||
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"I like clouds. I saw one shaped like a vagina once. " I think I saw that cloud too, perhaps it was the same clunge shaped cloud we were both spying... It all depends what the farthest cloud you can see is... | |||
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"I like clouds. I saw one shaped like a vagina once. I think I saw that cloud too, perhaps it was the same clunge shaped cloud we were both spying... It all depends what the farthest cloud you can see is... " I know it was a cum—u-less | |||
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"You can't see beyond the horizon " The horizon is the furthest distance of the land you can see due to the curvature of the earth. The clouds being much higher are much further away. | |||
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"I like clouds. I saw one shaped like a vagina once. I think I saw that cloud too, perhaps it was the same clunge shaped cloud we were both spying... It all depends what the farthest cloud you can see is... I know it was a cum—u-less " You've been waiting years to use that one... | |||
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"I like clouds. I saw one shaped like a vagina once. I think I saw that cloud too, perhaps it was the same clunge shaped cloud we were both spying... It all depends what the farthest cloud you can see is... I know it was a cum—u-less You've been waiting years to use that one... " Hook, line and sinker | |||
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"I dunno 17 miles maybe " Oh come on, I expected better than that.. Its just some simple trigonometry combined with Pythagorus... simple.. | |||
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"32 Km away" Please show your workings out... | |||
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"32 kilometres and that’s 30 add 2 " Actually, I had no idea at all. | |||
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"32 Km away Please show your workings out... " Asked Google nautical | |||
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"I’m learning Dutch " hoe vind je het? | |||
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"Roughly 223 miles?" Actually, ~226 miles, forgot the 3.2 | |||
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"4.2 Miles " Nope.. The sky that is visible is perpendicular to land not visible beyond the horizon. | |||
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"32 Km away Please show your workings out... Asked Google nautical " tut, tut, tut... caught cheating again ehh... | |||
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"You wouldn't be able to distinguish an individual cloud. Because the sky follows the same curvature as the Earth, the clouds would appear joined after about 3 miles." Its a vagina shaped cloud.. its very distinguished.. | |||
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"32 Km away Please show your workings out... Asked Google nautical tut, tut, tut... caught cheating again ehh..." Its called initiative | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon?" From your position to the furthest cloud is approx 143.3km or 89.04 miles assuming continuous cloud layer at 1 mile high and the perfection of God-like vision. BTW Mixing metres and miles in a calculation isn't a good idea. | |||
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"Its 142.71km to be exact" Have a tick, though my calculation gives 143.3 km. Very probably due to whether or not one takes average earth radius. | |||
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"Its 142.71km to be exact Have a tick, though my calculation gives 143.3 km. Very probably due to whether or not one takes average earth radius." Yeah, i was using 6371km for the radius. I think i rounded down converting the miles to km, so i think its probably just over 143km | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon?" What if it’s a clear day? | |||
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"You can't see beyond the horizon The horizon is the furthest distance of the land you can see due to the curvature of the earth. The clouds being much higher are much further away. " Are they far away, or are they just small? | |||
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"You wouldn't be able to distinguish an individual cloud. Because the sky follows the same curvature as the Earth, the clouds would appear joined after about 3 miles. Its a vagina shaped cloud.. its very distinguished.. " Ah, a cunninimbus. | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km" True if you ignore the curvature of the earth. In which case you have to use the distance to the horizon as the section. Work out the length of the segment Use sine rule to work out the distances from eye level to the horizon and from the horizon to the clouds. | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km" Is that taking into account the deflection of light due to gravitational lensing? | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km Is that taking into account the deflection of light due to gravitational lensing? " With something the mass of the earth it's not a huge effect | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km Is that taking into account the deflection of light due to gravitational lensing? With something the mass of the earth it's not a huge effect" I'm told it's theta = 4GM/rc2. Simples really isn't it! | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon?" You can't see beyond the horizon! | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km" Okay..Now This just fried my brain! Ditzy Maz strikes again xx | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon?" Trik question. You carnt see beyond the horizon | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon? Trik question. You carnt see beyond the horizon " The horizon is measured by visible land. The sky horizon is variable depending upon the cloud height. | |||
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"You can't see beyond the horizon The horizon is the furthest distance of the land you can see due to the curvature of the earth. The clouds being much higher are much further away. Are they far away, or are they just small?" Now now, _orticia.. Lets go through this one more time... The cows in the field are far away and this one here is very small.. Far away... very small.. do you get it now?... | |||
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"How many sweets did Johnny have again " None, Jane stole them all. | |||
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"I’ve had a word with Google and he says you cant see beyond 32km with the naked eye Hence his answer " You mean the moon, sun and assorted other stars are less than 32km away? | |||
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"The question is, if you stand at sea level looking out to sea, you are 1.8m tall, the distance you can see to the horizon is around 3.2 miles. The clouds are at a constant height of 1 mile, therefore how far away is the furthest cloud you can see beyond the horizon? Trik question. You carnt see beyond the horizon The horizon is measured by visible land. The sky horizon is variable depending upon the cloud height. " In that case I'm not Shor | |||
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"The cloud is a problem, it clouds the answer as it technically blocks light which the eye needs to see. Because you would be looking through an increasing density of water vapour as you looked further out to the sky, you would need to know the angle of the light source relative to your position. And the clarity of the space between your view point and the reflected light from the target cloud. Distance is not a problem if you are looking at a light source e.g. a distant star, but a reflected light source is much more complex. Also as I get older my book needs to get closer to the horizon to become readable which is bloody annoying when I have to walk over to turn the page " Sounds like you need to get large print versions of your reading material.. Or get a new pair of reading glasses, that pair of binoculars you use to read your books kept in the bedroom of the house opposite looks just a little suspicious.. | |||
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"I’ve had a word with Google and he says you cant see beyond 32km with the naked eye Hence his answer You mean the moon, sun and assorted other stars are less than 32km away? " That explains why it has been so warm the last few days.. | |||
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"The cloud is a problem, it clouds the answer as it technically blocks light which the eye needs to see. Because you would be looking through an increasing density of water vapour as you looked further out to the sky, you would need to know the angle of the light source relative to your position. And the clarity of the space between your view point and the reflected light from the target cloud. Distance is not a problem if you are looking at a light source e.g. a distant star, but a reflected light source is much more complex. Also as I get older my book needs to get closer to the horizon to become readable which is bloody annoying when I have to walk over to turn the page Sounds like you need to get large print versions of your reading material.. Or get a new pair of reading glasses, that pair of binoculars you use to read your books kept in the bedroom of the house opposite looks just a little suspicious.. " Drat, busted | |||
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"The cloud is a problem, it clouds the answer as it technically blocks light which the eye needs to see. Because you would be looking through an increasing density of water vapour as you looked further out to the sky, you would need to know the angle of the light source relative to your position. And the clarity of the space between your view point and the reflected light from the target cloud. Distance is not a problem if you are looking at a light source e.g. a distant star, but a reflected light source is much more complex. Also as I get older my book needs to get closer to the horizon to become readable which is bloody annoying when I have to walk over to turn the page Sounds like you need to get large print versions of your reading material.. Or get a new pair of reading glasses, that pair of binoculars you use to read your books kept in the bedroom of the house opposite looks just a little suspicious.. Drat, busted " Having your tallywhacker in your hand while "reading" was the real give away... | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km True if you ignore the curvature of the earth. In which case you have to use the distance to the horizon as the section. Work out the length of the segment Use sine rule to work out the distances from eye level to the horizon and from the horizon to the clouds. " The curvature of the earth is taken into account as we are using the radius for the calculation, we are assuming the earth is a perfect circle, which it probably isn’t, but otherwise the logic is sound. It would be easier to explain if I could draw a diagram. Just imagine a circle drawn in a piece of paper to represent a cross section of the earth, then draw a slightly larger circle around it to represent the cloud cover at a consistent distance from the surface. Draw a line from the centre of the circles straight up and slightly beyond the inner circle to represent someone standing 1.8m from the surface of the earth, then at a right angle from that point draw another line until it meets the outer circle, representing the cloud cover. Finally, draw another line from that point back to the centre of the circles. This gives us the right angled triangle to make the calculation and also taking the curvature of the earth into account. | |||
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"This is actually pretty simple if you think about it, you just need basic Pythagoras. a^2+b^2=c^2 with c being the hypotenuse Now imagine a cross section of the earth with you standing on top. The points of the triangle run from the centre of the earth to your eye, this being a right angle from eye to the clouds, and finally from the clouds back to the centre of the earth. The radius of the earth is approx. 6371km. For accurate calculation we need to convert everything to the same units, lets use km. radius=6371km height to eye= 0.0018km height to clouds= 1.609km ignore the distance to the horizon, this is a red herring. So, back to the formula, we have the hypotenuse and the other longest side so we need to rearrange it as follows to find the shortest distance: a^2=c^2-b^2 Now we just plug in the values a^2=(6371+1.609)^2-(6371+0.0018)^2 a^2=40610145.466881-40589663.93560324 a^2=20481.53127776 a=143.1137km True if you ignore the curvature of the earth. In which case you have to use the distance to the horizon as the section. Work out the length of the segment Use sine rule to work out the distances from eye level to the horizon and from the horizon to the clouds. The curvature of the earth is taken into account as we are using the radius for the calculation, we are assuming the earth is a perfect circle, which it probably isn’t, but otherwise the logic is sound. It would be easier to explain if I could draw a diagram. Just imagine a circle drawn in a piece of paper to represent a cross section of the earth, then draw a slightly larger circle around it to represent the cloud cover at a consistent distance from the surface. Draw a line from the centre of the circles straight up and slightly beyond the inner circle to represent someone standing 1.8m from the surface of the earth, then at a right angle from that point draw another line until it meets the outer circle, representing the cloud cover. Finally, draw another line from that point back to the centre of the circles. This gives us the right angled triangle to make the calculation and also taking the curvature of the earth into account." Ya what now xx | |||
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"Why draw the sight line at 90 degrees? Surely it should be drawn as a tangent?" I did this for ease of explanation assuming a straight line of sight, however you are correct. It can still be calculated, but it just makes things a lot more complicated and the difference it makes is negligible. The key to solving the problem is in the logic of visualising of a cross section of the earth and drawing an imaginary triangle from the earths centre to the points of interest. Every triangle has 6 pieces of information, the 3 lengths and the 3 angles, as long as we have 3 of these, we can work out the other 3 using the sine rule and cosine rule. As you pointed out, it wouldn't be a perfect 90 degree line of sight, but the logic for the calculation remains the same. The OP stated the horizon is visible 3.2 miles away, so we can use this to calculate the angle of the line of sight, and, in turn, the distance of the clouds. For this we can assume that the horizon (the farthest piece of land visible due to the curvature of the earth) is 3.2 miles from your eyeball (it was not specifically stated if it was from the eyeball or the feet), we must also assume that the eyeball is 1.8m from the ground, not the height of the top of the head. So now that we have the 3 lengths of the scalene triangle, we can calculate the angles using the sine and cosine rules, and then use the angle of the line of sight to calculate the distance to the clouds. Having done the calculations, the angle of the line of sight is actually 89.9568168 degrees, not 90 degrees, which ultimately means the cloud distance is 147.995998km instead of 143.1137km. I can show the workings if anyone is genuinely interested. | |||
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"I like clouds. I saw one shaped like a vagina once. " Hahahaha | |||
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"We were playing out in our car last week at the top of a hill, and that exact question came up, the answer we settled on was "shut up and keep sucking, who gives a fxxk how far that cloud is" ... hope that helps, it worked for us " best answer yet | |||
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"Why draw the sight line at 90 degrees? Surely it should be drawn as a tangent? I did this for ease of explanation assuming a straight line of sight, however you are correct. It can still be calculated, but it just makes things a lot more complicated and the difference it makes is negligible. The key to solving the problem is in the logic of visualising of a cross section of the earth and drawing an imaginary triangle from the earths centre to the points of interest. Every triangle has 6 pieces of information, the 3 lengths and the 3 angles, as long as we have 3 of these, we can work out the other 3 using the sine rule and cosine rule. As you pointed out, it wouldn't be a perfect 90 degree line of sight, but the logic for the calculation remains the same. The OP stated the horizon is visible 3.2 miles away, so we can use this to calculate the angle of the line of sight, and, in turn, the distance of the clouds. For this we can assume that the horizon (the farthest piece of land visible due to the curvature of the earth) is 3.2 miles from your eyeball (it was not specifically stated if it was from the eyeball or the feet), we must also assume that the eyeball is 1.8m from the ground, not the height of the top of the head. So now that we have the 3 lengths of the scalene triangle, we can calculate the angles using the sine and cosine rules, and then use the angle of the line of sight to calculate the distance to the clouds. Having done the calculations, the angle of the line of sight is actually 89.9568168 degrees, not 90 degrees, which ultimately means the cloud distance is 147.995998km instead of 143.1137km. I can show the workings if anyone is genuinely interested. " No thanks . My head will explode | |||
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