# Height h at which something d miles away is visible **URL:** \ **Category:** Factual Questions\ **Created:** [July 13, 2021, 3:43pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399 "2021-07-13T15:43:53Z")\ **Posts on this page:** 15\ **Page:** 1
**Author:** ![whc.03grady](https://avatars.discourse-cdn.com/v4/letter/w/a6a055/32.png) [@whc.03grady](https://boards.straightdope.com/u/whc.03grady)\ **Post date:** [July 13, 2021, 3:43pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/1 "2021-07-13T15:43:53Z")
Assume a sphere, sea level, blah blah. How tall does something have to be for me to see it above the horizon if it’s 70 miles away? All I can find online are distance to the horizon calculators…though maybe those would work if I could figure out how to frame the question within them. ---
**Author:** ![DesertDog](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/desertdog/32/11503_2.png) [@DesertDog](https://boards.straightdope.com/u/DesertDog)\ **Post date:** [July 13, 2021, 4:01pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/2 "2021-07-13T16:01:49Z")
Something over 3,000 feet (70mi distance, 5.5ft eye height) [Earth Curvature Calculator](https://www.omnicalculator.com/physics/earth-curvature) Flat earthers use calculators like this all the time in their proofs. ---
**Author:** ![mtnmatt](https://avatars.discourse-cdn.com/v4/letter/m/a3d4f5/32.png) [@mtnmatt](https://boards.straightdope.com/u/mtnmatt)\ **Post date:** [July 13, 2021, 4:17pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/3 "2021-07-13T16:17:02Z")
I’m not sure about the formulas, but on a very clear day, if you knew where to look, you could see the top floors of the WTC (1368ft) from Fairfield Beach in CT, which is about 49 miles away as the crow flies. ---
**Author:** ![Marvin\_the\_Martian](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/marvin_the_martian/32/2898_2.png) [@Marvin\_the\_Martian](https://boards.straightdope.com/u/Marvin_the_Martian)\ **Post date:** [July 13, 2021, 4:29pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/4 "2021-07-13T16:29:39Z")
The calculators don’t take into account refraction from the atmosphere (geodetic refraction) which allows you to see things that are beyond the mathematical horizon. ---
**Author:** ![DesertDog](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/desertdog/32/11503_2.png) [@DesertDog](https://boards.straightdope.com/u/DesertDog)\ **Post date:** [July 13, 2021, 4:42pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/5 "2021-07-13T16:42:19Z")
This is true, and a fact the FE proponents never talk about but I was assuming the OP included that in the assumptions. “Assume a spherical cow.” ---
**Author:** ![Marvin\_the\_Martian](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/marvin_the_martian/32/2898_2.png) [@Marvin\_the\_Martian](https://boards.straightdope.com/u/Marvin_the_Martian)\ **Post date:** [July 13, 2021, 4:49pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/6 "2021-07-13T16:49:04Z")
I actually meant to reply to mtnmatt to point out why there might be disagreement between the calculators and his observations (I didn’t actually check his numbers to see if there is disagreement). ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [July 13, 2021, 4:53pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/7 "2021-07-13T16:53:40Z")
They also miscalculate any difference in elevation. Sure, your eyesight level might be 6 feet, but you’re standing on a hill 450 feet taller than the object you’re looking at! ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 13, 2021, 5:09pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/8 "2021-07-13T17:09:54Z")
Suppose your eye is 1.70m above the ground. Ignoring temperature and pressure corrections and so on, the distance to the visible horizon as per navigation tables will be approximately 2.71 nautical miles. Now 70 “miles” is about 60 nautical miles, so the tall object is, let’s say, 58 nautical miles beyond your horizon, therefore it needs to be at least (58/2.08)^2 = 777m tall. The printed table does not actually go up that high, but it should give you the idea that you are not talking about something 5 m off the ground. ---
**Author:** ![DesertDog](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/desertdog/32/11503_2.png) [@DesertDog](https://boards.straightdope.com/u/DesertDog)\ **Post date:** [July 13, 2021, 5:10pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/9 "2021-07-13T17:10:41Z")
Also true. Even an additional 5 feet above the water surface reduces the minimum height to 2,900 feet. I was not meaning to imply FE proofs are valid. ---
**Author:** ![abcdefghij](https://avatars.discourse-cdn.com/v4/letter/a/43a26b/32.png) [@abcdefghij](https://boards.straightdope.com/u/abcdefghij)\ **Post date:** [July 13, 2021, 5:45pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/10 "2021-07-13T17:45:27Z")
The way I was taught to sink targets[\*] was was distance in nautical miles is 1.1 times square root of height of eye (in feet). So, h=\left(\frac{70\,\mathrm{mi}}{1.15\,\mathrm{mi/nm}\times1.1\,\mathrm{nm/ft^2}}\right)^2=3062\,\mathrm{ft}, which is what @DesertDog said. [\*] as in “there are two kinds of ships: submarines and targets” ---
**Author:** ![crowmanyclouds](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crowmanyclouds/32/19884_2.png) [@crowmanyclouds](https://boards.straightdope.com/u/crowmanyclouds)\ **Post date:** [July 13, 2021, 7:24pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/11 "2021-07-13T19:24:08Z")
> [@Marvin\_the\_Martian](#): > > The calculators don’t take into account refraction from the atmosphere (geodetic refraction) which allows you to see things that are beyond the mathematical horizon. [Walter Bislin’s](http://walter.bislins.ch/bloge/index.asp?page=advanced+earth+curvature+calculator) does, you can ever change the refraction rate. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [July 14, 2021, 7:21pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/12 "2021-07-14T19:21:56Z")
> [@whc.03grady](#): > > All I can find online are distance to the horizon calculators…though maybe those would work if I could figure out how to frame the question within them. You can use those, just add the distances. If your eye is 6 ft off the water surface, you get one number for distance to the horizon. Then you can find the distance to the horizon for some distant object. To get the total distance from you to that object, just add together the two distances to the horizon. To answer your specific question, you’d do one of these steps backwards: find the distance to the horizon for your eyes. Let’s say that’s 3 miles. For something 70 miles from you, you’d then find the height for an object with a distance to the horizon of 67 miles. You can see why this is by drawing a simple diagram. Then line from your eye to the horizon, and the line from the top of the distant object to the horizon (towards you) are the same line. ---
**Author:** ![Keeve](https://avatars.discourse-cdn.com/v4/letter/k/f07891/32.png) [@Keeve](https://boards.straightdope.com/u/Keeve)\ **Post date:** [July 14, 2021, 9:03pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/13 "2021-07-14T21:03:27Z")
> [@crowmanyclouds](#): > > [Walter Bislin’s](http://walter.bislins.ch/bloge/index.asp?page=advanced+earth+curvature+calculator) does, you can ever change the refraction rate. That’s helpful, because the refraction changes with the current weather. ---
**Author:** ![iamthewalrus\_3](https://avatars.discourse-cdn.com/v4/letter/i/258eb7/32.png) [@iamthewalrus\_3](https://boards.straightdope.com/u/iamthewalrus_3)\ **Post date:** [July 14, 2021, 9:36pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/14 "2021-07-14T21:36:49Z")
You can frame this question as two distance to horizon questions, which are just Pythagorean theorem. There’s one right triangle with short edges equal to the radius of the earth _r_, and distance to the observer _d1_, hypotenuse _r_ + _height of observer’s eyes_. There’s another right triangle with short edges _r_, distance to the tall thing _d2_, and hypotenuse _r_ + _h_. _d_ = _d1_ + _d2_ ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 14, 2021, 11:37pm UTC](https://boards.straightdope.com/t/height-h-at-which-something-d-miles-away-is-visible/946399/15 "2021-07-14T23:37:34Z")
> [@CurtC](#): > > You can use those, just add the distances. Except that he’s solving the inverse problem to what most of those calculators do. He knows how far away it is, and wants to calculate the height from that. Of course, you can still do that with such a calculator, by plugging in heights until you find one that gets you the distance you want (possibly using the method of bisection, or Newton’s method, to speed things up). But it’s clunkier than just typing in one number and pressing a button.