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### Earth's Curvature

```
Date: 01/27/97 at 17:20:57
From: Neha
Subject: Earth's curvature

I am interested in knowing the degree of curvature of the earth's
surface.  I am attempting to find a method which would allow me to
conceptualize and calculate the making of such a depiction on paper.

That is to say, if the earth curves from point A to point B, points
exactly 10 kilometers apart, if one were to draw a straight line
betwen the two points, then WHAT WOULD BE THE HEIGHT of the highest
point the curvature AB would be above the line AB?  Moreover, if the
point AB were now to be 20 kilometers apart, would the height be 2
times the height calculated earlier or would it be something
altogether different?

Thank you for your kind consideration and aid.  This young artist
sincerely and surely appreciates it!
```

```
Date: 01/29/97 at 12:01:59
From: Doctor Lorenzo
Subject: Re: Earth's curvature

Hi Neha -

The first thing to realize is that this problem is about circles in
the plane, not about spheres in 3-dimensional space.  You are only
interested in comparing a curve and a straight line, and they all lie
in the same plane. A plane through a sphere yields a circle.  So we
just have to understand how a circle pulls away from a straight line.

Consider a circle of radius R, tangent to the x-axis at the origin and
the center on the negative y-axis.  (I set it up this way in order to
simplify the calculation of the height which now will go straight up
the y-axis from the midpoint of line AB to the origin.) The equation
for such a circle is:

x^2 + (y+R)^2 = R^2

Equivalently, x^2 + y^2 +2Ry = 0

Equivalently, y = - (x^2 + y^2)/2R

[Note: The notation "x^2" means "x squared"]

Near the origin y is small, and y^2 is smaller still, so this can be
approximated by:

y = - x^2 / 2R

If 2 points are separated by a distance L, they might be point A at
x = -L/2 and point B at x = +L/2.  At both A and B we have:

y = - L^2 / 8R

That is, while the curve from A to B goes through the origin, the
straight line from A to B cuts a distance L^2/8R below the origin.
Notice that doubling L (say, from 10 miles to 20) means quadrupling
this distance.

As for the curvature of the actual earth, you just have to plug in,
for R, the radius of the earth, which is about 4000 miles.  So, if
L = 10 miles, then:

y = (100/32,000) miles (about 16 feet)

You can see this effect in real life.  If you look across a 10-mile
wide lake at the other shore, you can't see the low buildings on the
other side! To see a cabin 16 feet high, you would have to climb a
ladder so that your eyes are 16 feet above the ground (to be exact you
would take into account the effect of the curvature on the angles of
the building and your viewing station make relative to the "y-axis",
but this effect is negligible at this distance).

This is dramatic in Lake Tahoe, where, from the north shore of the
lake you can see the tall casinos of Stateline, Nevada, but not the
low buildings of South Lake Tahoe, California.  (Lake Tahoe is also
more than 10 miles from top to bottom).

By the way, the L^2/8R answer is only an approximation, since we
ignored the y^2 term in the equation of the circle, and ignored the
difference between the straight-line distance between A and B and the
distance along the curve.  These corrections are negligible, being
approximately of size L^4/R^3. If L = 10 miles and R = 4000 miles,
these corrections add up to less than 1/1000 of an inch!

-Doctor Lorenzo,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 12/07/2004 at 06:13:46
From: Martin

Hi -

Your reply to this question assures us that the Math is the same for
all the calculations and the only things that change are the radius
of the circle and the distance from a to b.

I have drawn a circle with a 60mm radius and tried out the
calculations.  Where "L" (distance between points a and b) is very
short, say 40mm, then the result "Y" is virtually accurate.

However, as you lengthen "L" to say 115mm the result "Y" becomes
wholly inaccurate.  Instead of a measured result of 42mm the
calculation result is 27mm.

Can you shed any light on this for me please?  Am I missing
something very simple?
```

```
Date: 12/07/2004 at 09:37:44
From: Doctor Vogler
Subject: Re: Problem with your archive

Hi Martin,

Thanks for writing to Dr Math.  The last paragraph in the reply reads:

By the way, the L^2/8R answer is only an approximation, since we
ignored the y^2 term in the equation of the circle, and ignored the
difference between the straight-line distance between A and B and the
distance along the curve.  These corrections are negligible, being
approximately of size L^4/R^3.  If L = 10 miles and R = 4000 miles,
these corrections add up to less than 1/1000 of an inch!

So essentially he is saying that when L is small compared to R, then
the approximation is very good.  In your example, L is not small
compared to R; in fact, L is bigger than R.

So let's get a more exact result.  You'll recall that the
approximation was in neglecting the y^2 term in

x^2 + y^2 + 2Ry = 0.

So what happens if we DON'T neglect the y^2 term?  Then when we
substitute x = L/2, we get

y^2 + (2R)y + (L/2)^2 = 0

and we solve for y using the quadratic formula:

y = -R + (1/2)sqrt(4R^2 - L^2)

and

y = -R - (1/2)sqrt(4R^2 - L^2),

and then the height to the x axis would be

-y = R - (1/2)sqrt(4R^2 - L^2)

or

-y = R + (1/2)sqrt(4R^2 - L^2),

If you look on the circle, you notice that one of these answers is
measuring the height where the curved line around the circle takes the
shorter way around the circle (that's the first answer) and the other
gives the answer if the curved line takes the long way around the circle.

In the archived answer, he was looking for the short way, and you'll
find that if L is small compared to R then the square root can be
approximated by two terms of the Taylor series,

sqrt(4R^2 - L^2) is about 2R(1 - (1/2)(L/2R)^2)

where the difference will be roughly the size of the third term of the
Taylor series, which is (1/8)(L/2R)^4.  That approximation gives the
same answer as recorded in the archive, L^2/8R.  But if (L/2R)^4 is
not a small number, then this is certainly not a good approximation,
and so you should use the exact answer,

R - (1/2)sqrt(4R^2 - L^2).

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/08/2004 at 04:09:23
From: Martin
Subject: Thank you (Problem at mathforum.org/library/drmath/view/54904.html)

Thank you very much.  Works a treat.  Faith in math restored.
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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