## Introduction

I frequently hear people make statements that somehow do not seem right. Today I heard a good example. During a discussion of laser levels, the topic of required accuracy came up. I heard a contractor make the following statement:

Who cares if the laser level is accurate within an 1/16th of inch at 100 feet? The Earth curves away from a horizontal line by 1/8th of inch for every 100 feet of horizontal distance. The Earth's curvature will swamp out your instrument error in less than 100 feet.

The statement about the curvature of Earth got me thinking. How much does the Earth's surface deviate from a horizontal line over a distance of 100 feet? The contractor's number intuitively seemed wrong because the Earth is round and the deviation from horizontal should be a function of distance. A little math will give me the answer. For consistency's sake, I will perform all computations in US customary units.

## Analysis

Figure 1 illustrates the situation and contains the derivation of both an exact and a approximate solution. The triangle formed by *x*, *R + δ*, and *R* is a right triangle, which means that the Pythagorean theorem can be used to produce an exact solution. In addition, a simple approximation for *δ* is also developed assuming *R* >> *x* and using a linear approximation for the square root. In Appendix A, I give examples of the computations in Mathcad.

Given the situation shown in Figure 1, we can compute the deviation from horizontal as follows.

Eq. 1 |

where

*R*is the radius of the Earth (3963.2 miles)*x*is the horizontal distance of interest (100 ft)

## Conclusion

The contractor had stated that the curvature of the Earth causes level to deviate from horizontal by an 1/8th of an inch (125 thousandths of inch) for 100 feet of horizontal distance. The actual deviation is ~2.9 thousandths of an inch for 100 feet of horizontal distance, which is almost 45 times less than the contractor claimed. So it is meaningful to buy a laser level that is accurate to 1/16th of an inch over 100 feet, i.e. the laser level error is not swamped by the curvature of the Earth.

Why did the contractor make the claim that the Earth's curvature is 1/8th inch over 100 feet? He made a simple mistake. He did not understand that the deviation from horizontal for short distances is given by a square-law relationship, shown in Equation 2. In Equation 2, I include an approximation that is only valid when *R* is much greater than *x*, which is true in typical construction problems.

Eq. 2 |

If we use Equation 1 or Equation 2 to calculate the deviation from horizontal at 1 mile, we get 8 inches. This value is quoted in a number of references on surveying (e.g. here is one, here is another, yet another). What the contractor did was erroneously assume that the deviation varied linearly with distance, which would mean that a deviation of 8 inches at 1 mile is equivalent to an 1/8th of inch at 100 feet.

For those of you who may be interested in the related question of the error in horizontal distances caused by living on a spherical planet, see this blog post.

Aside: Here is an interesting discussion that references this web page.

## Appendix A: Computation Examples

Figure 2 shows a few computation examples. Normally, I let Mathcad do the unit conversion, but I do show one example with explicit unit conversion.

hello, I was confused by your quotient because you state the distance in kilometres and the elevation in feet, and then at the end you quote the distance in miles and the elevation in inches. this is crazy and confusing! Using 4 units of measurement with only one equation, when there should only be one. it doesn't say anything about the conversion in the equation, so I figure it could be metric, then when I use metric in the calculation, the result is different from what you say. I will actually have to look on another website! for the solution! Cheers! happy Christmas!

Dimensional analysis can help you keep things straight. I have added a small appendix that illustrates the calculations. Normally, I let my computer algebra system do the unit conversions for me, so I never actually worry about them. In the appendix, I illustrate how to perform the unit conversion. I hope that helps.

Merry Christmas.

Mathscinotes

I like your web page but I have a question. Since it's been over 40 years since I took algebra could you explain equation 2? When I use smaller numbers, like 3 for x, 4 for r I get different answers between the first step & the second step. 1 for the first step & 1.25 for the second step. Thanks

My article was not clear on the approximation and I have added some explanatory text. The approximation is only valid when R is much much bigger than x. This is true when R is the radius of the Earth and x is a typical construction distance. The approximation fails when when R and x are similar in size.

Hopefully, my clarification in the text will help. I am so used to doing these sorts of approximations that I do not even notice when I do them -- very bad habit.

Mathscinotes

Hi, Thanks for your calculation for the vertical error. I am however interested in figuring out how much the distance between two points vary with curvature of earth (as opposed to just using trigonometry). Could you help me in figuring this calculation out? Thanks

Hi Hanna,

I would like to try to help, but I do not completely understand your question. Are you trying to determine the distance between two points on the Earth's surface? If so, I can help. Generally distances on the Earth's surface are expressed in terms of the great circle distance. There are numerous web sites that address this calculation, like this one. If you are interested in a proof of the formula used on this web page, see this reference. If you decide that you need an Excel version of the calculation and are having trouble putting it together, I can pull something together in less than a minute.

If you have a different question, just give me a bit more detailed version of your question and I will try to help.

mathscinotes

Hi. For some reason I never saw your reply. Thanks for getting back to me. I am observing animals from a 93-96m high sea cliff (tidal variation) with a theodolite. I get angles from reference point that allow me to calculate the (diagonal) distance of the animal in the water from the theodolite on the cliff top. As I know the height at that moment, I can easily calculate the horizontal distance from the theodolite to the animal. However I would like to know what is the actual distance of the animal from the foot of the cliff, if the curvature of the earth is taken into account. I have tried using the moveable typescript calculations but am not sure if I can do it correctly. Assuming the angle from is 45degrees

Hi Hanna,

Let's take a quick look and see if I am thinking of your problem correctly. To keep things very simple, let's assume that your theodolite is 100 m above the ocean and you are observing something in the water at a 45 °C angle. Using "flat Earth" methods, your subject is 100 m away -- correct? Let's work the same problem and determine the arc length of your subject from a point directly below your theodolite.

Here is my drawing that illustrates this situation.

I can convert this situation into a geometry problem as shown in the following figure.

I can analyze this geometry as shown below.

As you would expect, the arc distance is not much different than the distance you would compute using simple angles.

Is this what you were looking for? I can put it into Excel if you want to play with it.

Mathscinotes

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votre blog

Hello:

I am trying to check and see if my math is correct. I found your article and wanted to see if I was on the right track.

I am trying to figure out the curvature of the earth based on the distance...

Thanks

James

Miles Squared X 8 Inches= Inches Of Curvature

6 Miles X 6 (Squared) = 36 x 8 = 288 Inches

288 / 12 = 24 Feet

Miles Curvature Drop

1 8 Inches

2 32 Inches

3 6 Feet

4 10 Feet

5 16 Feet

6 24 Feet

7 32 Feet

8 42 Feet

9 54 Feet

10 66 Feet

20 266 Feet

30 600 Feet

40 1,066 Feet

50 1,666 Feet

60 2,400 Feet

70 3,266 Feet

80 4,266 Feet

90 5,400 Feet

100 6,666 Feet

120 9,600 Feet

Close, it's slightly more than 8" per mile² so as you add more and more miles your answer drifts from the actual value (assuming a prefect sphere). Why not just use the exact equation instead of an approximate though?

Beyond that - there are a couple of very serious issues with this that people often fall into...

#1 the largest effect that makes this table almost COMPLETELY irrelevant is that it doesn't take into account the VERY SIGNIFICANT contribution from the refraction of light. When talking about 100 miles over the ocean this table will be off by DOZENS OF MILES in terms of where the ACTUAL APPARENT HORIZON appears. Thanks to refraction you can see much, much further "around" the ocean than most people seem to understand. The net effect of refraction is to make the appear a fraction larger around than it actually is. The *nominal* value for this is 1/7th of the curvature of Earth -- So if you want to get an approximate value for refraction you can use R*(7/6) to adjust for refraction. However, the actual value can be significantly GREATER than 1/7 - especially in sea level, warm, humid conditions like you expect to find around tropical islands. Try the same trick up around Greenland on a cold, dry day at higher altitudes and you'll find a significantly smaller effect.

The refraction of light can vary from very slight effects to the extreme end of COMPLETELY BENDING LIGHT AROUND THE GLOBE -- in fact, when you view a mirage the refraction is so extremely the light is bending upwards! The refraction of light in humid, dense air -- especially right above a body of water -- is very severe. The exact effect IS ALMOST IMPOSSIBLE TO CALCULATE because you almost never know, with sufficient precision, the exact structure of every molecule of air between you and the object viewed -- especially at a great distance. However, you can get a *pretty good* approximate if you control for the primary factors: frequency of the light, the density of the atmosphere, the gradient of density, the humidity, temperature, and CO2 content of the atmosphere.

If you want to account for all the conditions to the best of our ability then what you want to do is look up Ciddor and Edlen - these formulas are EXTREMELY complex - so good luck with that! But, thanks to our tax dollars, there is a web-based calculator! http://emtoolbox.nist.gov/Wavelength/Ciddor.asp

But the results are only as good as the input data -- if you make up data (oh, it was probably 30C on that day) then you get the wrong results.

What we do know for certain is that the effects of refraction, especially over the ocean, often allow you to see FAR FURTHER around the globe than you possibly could without refraction. Failing to account for this refraction is EXACTLY the error made in the https://en.wikipedia.org/wiki/Bedford_Level_experiment -- the view at 8" over the water was VASTLY different from the few a few feet higher. It actually LOOKED flat at 8" elevation - that's how strong the refraction was due to humidity differences.

And people were RIGHT to question this -- "hey it REALLY LOOKS flat" but the error was in failing to accept new information that clearly disproved the flat hypothesis.

I've reviewed DOZENS of images where people are ASTONISHED that you can see a distant island where the NAIVE equation says you shouldn't be able to see it -- in every case when you account for even AVERAGE refraction you get a curvature value that is RIGHT where it should be.

#2 the Earth is SLIGHTLY oblate, not perfectly sphere so actual values would be a TINY margin off from these values. You need to use a model for the Shape of the Earth that takes this into account if you want to get more exact values and those values will ONLY be accurate for a specific latitude, longitude, and viewing angle & distance (and doing so requires integration over the surface).

HOWEVER -- This is NOT a hugely significant effect. Why?

Well, the Equatorial radius is 6,378.1 km while the Polar radius is 6,356.8 km -- that's not a huge distance. In fact, in MOST pictures of the earth you are talking about A FEW PIXELS of difference - that's less than the error for most resolutions. The exception for this is in extremely high resolution images (11000x11000) such as from the Himawari 8 (Japanese weather sat) where you can see the Earth is about 38 pixels wider than it is high (about 1/100" on my screen). I cannot SEE the difference from a sphere because this is such a SMALL % difference, but I CAN measure the exact pixels and also calculated the expected values and they are in agreement to within the precision of the data I have.

#3 Even once you take the above into account you STILL haven't taken into account the shape of the landmasses and the Tides, which will affect the shape of the water you are viewing over.

#4 people forget they are viewing most of this slope almost edge-on, this has so many distorting effects on viewing I cannot even list them all.

But what I can tell you is that the culmination of everything is that YOU CANNOT visually discern the curvature of the earth below about 30,000 feet and even then you need a very wide angle of view and, to make matters worse, ALL LENSES, including your eyes, introduce barrel distortions and are unreliable unless you have the EXACT equations for that lens and know the exact lat, long, altitude, & angle.

This error goes for people posting images of rounded Earths as well -- MOST of the curvature you see in ANY picture below the ISS is barrel distortion. Even at very high altitudes you are only seeing a few hundreds of miles of a globe that is almost 8000 MILES across. Even 100,000 feet is just BARELY off the surface of this monster. People seem to have no idea of the scale of things.

41,850,000 feet wide and you are 100,000 feet up? Come on!

This is the best article I have seen on that side of the question.

Visually discerning the curvature of the Earth

http://thulescientific.com/Lynch%20Curvature%202008.pdf

Go here if you want to see what it REALLY looks like (from 35,800 km! up)

http://himawari8.nict.go.jp/

This beast takes 11000x11000 images in 16 DIFFERENT frequency bands (which includes visual, near-IR, and IR) and gives astonishing clarity.

You can watch a movie of an entire month of data (~4000 images)

I am taking big issue with this article.

Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).

I hate imperial so please allow me to convert it to metric.

0.07366 mm = 30.480m

Multiply all of this up by 1000 =

73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?

Those that come up with 8inches a mile are much closer to the truth.

Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.

I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???

Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.

If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.

Thank you. Very nicely put.

mathscinotes

So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?

This is a good question and it all comes down to how you define "swamped". Let's assume that a measurement becomes useless when the error of the instrument equals the curvature of the Earth. My laser level has an error of 1/8 inch per 100 yards (i.e. an angular error). With respect to curvature, this means that the instrument is useful out to about 484 yards. The calculations are as shown below.

The earth is flat and if you would do you maths and just natural obervations you would feel that it is not spinning and always raise to eye level no matter how high you go. If you feel it spinning please go and see your doctor.:-)

Thank you Juan,

finally the correct answer! Whether you use 8" per mile squared or 6.5 " per mile squared... you still shouldn't see the statue of liberty from 60 miles, or the NYC and Philly skylines from 60 miles....

Use your heads people, test for yourself! If something disappears "over the horizon" (ship, for example) then how come you can see it again if you use binoculars? Impossible if the earth were really curved!

tragedy of science – the slaying of a beautiful hypothesis by an ugly fact.

Huxley

Just a matter of time before a flat-Earther wandered in.

Sailboat going gradually beyond and below the horizon: https://www.youtube.com/watch?v=7nUFLLUahSI

Large schooner disappears behind the horizon -- from the hull upward: https://www.youtube.com/watch?t=35&v=dV0h68YU0iQ

High and low observation of off-shore island: https://www.youtube.com/watch?v=bco_p4V7-QU

Clear demonstration of Earth's curvature on the Thames River, (skip to 5:00 to get to the beef):

https://www.youtube.com/watch?v=jRdgnj17FHQ

Base of buildings blocked from view by the water's convex curvature:

https://en.wikipedia.org/wiki/Horizon#/media/File:Horizon,_Valencia_%28Spain%29.JPG

Your reference material is excellent! Thank you very much.

mathscinotes

Malvarrosa Beach, Valencia

https://commons.wikimedia.org/wiki/File:Horizon,_Valencia_(Spain).JPG

1. Focal length 72mm of camera

2. People in the sea in reasonable in focus

3. Sea level ???? who knows

The focal length alone can create this effect.

Height above sea level looks to be about 6 foot.

Good illusion, well done.

Juan - Go to a playground. Get on a merry go round. Have someone you trust give you a push so gentle that the playground equipment spins at the rate of one rotation per 24 hours. If you feel the movement, please do go see the Dr. Then revisit your comments and report back, if you would.

You always find plumb to the center of the earth at whatever point your at on earth. All so let's say you were building a skyway from newYork to France. you would always stay x amount of feet above sea level bieng your reference benchmarks would always be right in front of you the (ocean surface) it would curve naturally with the earth. Railroads, roads are biult the same way. The benchmarks given to builders are close enough together so they don't have to account for curvature. Example : you could not use a fixed benchmark above sea level as a reference for grade at one point and the keep using that benchmark for the length of the railroad hypotheticly. Even if you had a laser thet was flawless and high enough off of grade to not be unstructed in anyway. your grades would start to become too high.

pseudoscience at its best! you always seem to have an excuse.......

contrary to nature ...

+water does not 'bunch up' at one end of the pond or lake..

+grammar school observations show that water is AlwAYs self leveling

+gyroscopes do NOT 'rotate' with the imaginary 'spin' of the plane(t) any more than 'day becomes night' 6 months from now as we rotate around a sun that is allegedly 93 million miles away

+infrared light does not 'refract' AND the Military IR Scopes can see 40 plus miles... but the plane(t) curve is 8 in per mile squared... ??

Could you please figure the curvature using 1 mile. Math has never been my forte

This link lists the curvature at 1 mile as 8 inches. If you need to actually need to see calculations, just send me another note.

mathscinotes

I recommend this formula for "drop off at distance d" assuming a perfect sphere (note: assumes d < R)

√[R² + d²] - R

For your case, you can plus this right into WolframAlpha:

√[(Earth average radius)² + (1 mile)²] - (Earth average radius) to inches

http://www.wolframalpha.com/input/?i=%E2%88%9A%5B(Earth+average+radius)%C2%B2+%2B+(1+mile)%C2%B2%5D+-+(Earth+average+radius)+to+inches

Which is ~8.0065 inches -- so DO NOT trust 8" d² for larger values, you gotta use more precise numbers or your error accumulates.

To give a numerical example, using mean radius of Earth in this case, still at 1 mile:

√[(3958.7613)² + (1)²] - (3958.7613) ~ 0.0001263 miles ~ 8.002368"

So as you can see the exact value you use for radius of Earth matters and that varies based on where you actually are - the Earth isn't a perfect sphere, it's an oblate spheroid: https://en.wikipedia.org/wiki/Earth_radius

So at 100 miles that is 1.263 miles drop off.

This doesn't take into account significant affects of refraction of light for visibility however. I've posted in another comment about that:

http://mathscinotes.com/2010/11/straight-level-and-the-curvature-of-the-earth/#comment-28337

Greetings,

I am a navigator and I use geographic ranging to calculate line of sight from an object with a known height to its line of sight horizon point. I use 1.169 times the square root of the height. Simply example: 1.169 x Square root of 100 equals 11.69 nautical miles to the point of horizon,

As I read Nathaniel Bowditch, the generally accepted master of navigational mathematics, I find he seems to use 1.146 instead of 1.169.

I am not sure why the accepted math taught today says to use 1.169.

SO I did some thinking.....Could the variation between Bowditch's math and the modern math have anything to do with the 14 mile difference between the polar radius and the equatorial radius?

Lastly, how did we come to use 1.169?

Thank you very much for any insight you would provide

Respectfully

Chris

Hi Captain,

It is quite likely that both Bowditch and the modern result are correct – it depends on the air conditions! First, I have looked at how to compute the distance to the horizon on this post, which uses metric units. I will convert to your units for this response.

If I ignore refraction, the distance to the horizon is given by , where d is the horizon distance in nautical miles and h is the height of the observer's eye in ft. To add refraction, I need to make an assumption on the rate of air temperature change with altitude – a parameter known as the lapse rate. If I assume 11.3°C/km (common), the horizon distance is given by . If I assume another lapse rate, I can get 1.169 for a leading coefficient. Yet another lapsre rate and I get 1.141 for a leading coefficient.

mathscinotes

Why is the drop measured from an angle from the center of the Earth rather than straight down from the horizontal/tangent(x)? For example, if I go ~4000 miles along x, I should show a ~4000 mile drop down from x. There seems to be a separate equation that perhaps doesn't use the Pythagorean theorem--can you show it? And if so, as 1) drop per movement along the horizontal and 2) drop per movement along the circumference.

Analysis, such Machinations, sheesh.

Try this approach with the given right triangle.

R^2 + X^2 = (R+delta)^2 – where R = 3963.2 mi. and X = 0.0185 mi. (or 100 ft)

15,706,954.24 + .00034 = 15,706,954.24034

√15,706,954.24034 = 3963.2 mi.

In other words, 100 feet will not demonstrate a visually noticeable difference.

And if you want to know delta (though why you’d want to know it I’m not sure as it is not necessary to the solution) it is √(15,706,954.24 - .00034) miles.

Or you could consider the distances used in the Bedford Level Experiment (https://en.wikipedia.org/wiki/Bedford_Level_experiment), in which Alfred Russell Wallace, co-discoverer of Natural Selection, participated.

Ooops, sorry on the length of delta. Should read √(15,706,954.24034) - 3963.2 miles.