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# The Global Grid Help (page 3)

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By — McGraw-Hill Professional
Updated on Oct 3, 2011

## Distance Per Unit Latitude

As measured along any meridian (that is, in a north-south direction), the distance d lat-deg per degree of latitude on the earth’s surface is always the same. It can be calculated by dividing the circumference of the earth by 360. If d lat-deg is expressed in kilometers, then:

d lat-deg = (4.003 × 10 4 )/360 = 111.2

The distance d lat-min per arc minute of latitude (in kilometers) can be obtained by dividing this figure by exactly 60:

d lat-min = 111.2/60.00 = 1.853

The distance d lat-sec per arc second of latitude (in kilometers) is obtained by dividing by exactly 60 once again:

d lat-sec = 1.853/60.00 = 0.03088

This might be better stated as d lat-sec = 30.88 meters. That’s a little more than the distance between home plate and first base on a major league baseball field.

## Distance Per Unit Longitude

As measured along the equator, the distances d lon-deg (distance per degree of longitude), d lon-min (distance per arc minute of longitude), and d lon-sec (distance per arc second of longitude), in kilometers, can be found according to the same formulas as those for the distance per unit latitude. That is:

d lon-deg = (4.003 × 10 4 )/360 = 111.2

d lon-min = 111.2/60.00 = 1.853

d lon-sec = 1.853/60.00 = 0.03088

These formulas do not work when the east-west distance between any two particular meridians is determined along a parallel other than the equator. In order to determine those distances, the above values must be multiplied by the cosine of the latitude θ at which the measurement is made. Thus, the formulas are modified into the following:

d lon-deg = 111.2 cos θ

d lon-min = 1.853 cos θ

d lon-sec = 0.03088 cos θ

The last formula can be modified for d lon-sec in meters, as follows:

d lon-sec = 30.88 cos θ

## The Global Grid Practice Problems

#### Practice 1

Imagine that a certain large warehouse, with a square floor measuring 100 meters on a side, is built in a community at 60° 0′ 0″; north latitude. Suppose that the warehouse is oriented “kitty-corner” to the points of the compass, so its sides run northeast-by-southwest and northwest-by-southeast. What is the difference in longitude, expressed in seconds of arc, between the west corner and the east corner of the warehouse?

#### Solution 1

The situation is diagrammed in Fig. 11-3. Let be the difference in longitude between the west and east corners of the warehouse. (The Δ symbol in this context is an uppercase Greek letter delta, which means “the difference in”; it’s not the symbol for a geometric triangle.)

Fig. 11–3. Illustration for Practice 1 and 2.

First, we must find the distance in meters between corners of the warehouse. This is equal to 100 × 2 1/2 , or approximately 141.4, meters. Now let’s find out how many meters there are per arc second at 60° 0′ 0″ north latitude:

In order to obtain , the number of arc seconds of longitude between the east and west corners of the warehouse, we divide 141.4 meters by 15.44 meters per arc second, obtaining:

We round off to three significant figures because that is the extent of the accuracy of our input data (100 meters along each edge of the warehouse). If we want to express this longitude difference in degrees, minutes, and seconds, we write:

#### Practice 2

What is the difference in latitude, expressed in seconds of arc, between the north and the south corners of the warehouse described above?

#### Solution 2

We already know that the distance between corners of the warehouse is 141.4 meters. We also know that there are 30.88 meters of distance per arc second, as measured in a north-south direction, at any latitude. Let Δ θ wh be the difference in latitude between the north and the south corners. We divide 141.4 meters by 30.88 meters per arc second, obtaining:

Again, we round off to three significant figures, because that is the extent of the accuracy of our input data (100 meters along each edge of the warehouse). If we want to express this longitude difference in degrees, minutes, and seconds, we write:

Practice problems for these concepts can be found at: Global Trigonometry Practice Test

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