Learning Objectives

- Use sigma (summation) notation to calculate sums and powers of integers.
- Use the sum of rectangular areas to approximate the area under a curve.
- Use Riemann sums to approximate area.

**Archimedes** was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape. He used a process that has come to be known as the **method of exhaustion**, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. These areas are then summed to approximate the area of the curved region.

In this section, we develop techniques to approximate the area between a curve, defined by a function (f(x),) and the x-axis on a closed interval ([a,b].) Like Archimedes, we first approximate the area under the curve using shapes of known area (namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking a limit allows us to calculate the exact area under the curve.

Let’s start by introducing some notation to make the calculations easier. We then consider the case when (f(x)) is continuous and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more general cases.

## Sigma (Summation) Notation

As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called **sigma notation** (also known as **summation notation**). The Greek capital letter (Σ), sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

[1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20.]

We could probably skip writing a couple of terms and write

[1+2+3+4+⋯+19+20,]

which is better, but still cumbersome. With sigma notation, we write this sum as

[sum_{i=1}^{20}i]

which is much more compact. Typically, sigma notation is presented in the form

[sum_{i=1}^{n}a_i]

where (a_i) describes the terms to be added, and the (i) is called the (index). Each term is evaluated, then we sum all the values, beginning with the value when (i=1) and ending with the value when (i=n.) For example, an expression like (displaystyle sum_{i=2}^{7}s_i) is interpreted as (s_2+s_3+s_4+s_5+s_6+s_7). Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a **dummy variable**. We can use any letter we like for the index. Typically, mathematicians use (i, ,j, ,k, ,m), and (n) for indices.

Let’s try a couple of examples of using sigma notation.

Example (PageIndex{1}): Using Sigma Notation

- Write in sigma notation and evaluate the sum of terms (3^i) for (i=1,2,3,4,5.)
- Write the sum in sigma notation:

[1+dfrac{1}{4}+dfrac{1}{9}+dfrac{1}{16}+dfrac{1}{25}. onumber]

**Solution**

- Write [sum_{i=1}^{5}3^i=3+3^2+3^3+3^4+3^5=363. onumber]
- The denominator of each term is a perfect square. Using sigma notation, this sum can be written as (displaystyle sum_{i=1}^5dfrac{1}{i^2}).

Exercise (PageIndex{1})

Write in sigma notation and evaluate the sum of terms (2^i) for (i=3,4,5,6.)

**Hint**Use the solving steps in Example (PageIndex{1}) as a guide.

**Answer**(displaystyle sum_{i=3}^{6}2^i=2^3+2^4+2^5+2^6=120)

The properties associated with the summation process are given in the following rule.

Rule: Properties of Sigma Notation

Let (a_1,a_2,…,a_n) and (b_1,b_2,…,b_n) represent two sequences of terms and let (c) be a constant. The following properties hold for all positive integers (n) and for integers (m), with (1≤m≤n.)

- (displaystyle sum_{i=1}^n c=nc)
- (displaystyle sum_{i=1}^n ca_i=csum_{i=1}^na_i)
- (displaystyle sum_{i=1}^n(a_i+b_i)=sum_{i=1}^na_i+sum_{i=1}^nb_i)
- (displaystyle sum_{i=1}^n(a_i−b_i)=sum_{i=1}^na_i−sum_{i=1}^nb_i)
- (displaystyle sum_{i=1}^na_i=sum_{i=1}^ma_i+sum_{i=m+1}^na_i)

Proof

We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.

2. We have

[sum_{i=1}^nca_i=ca_1+ca_2+ca_3+⋯+ca_n=c(a_1+a_2+a_3+⋯+a_n)=csum_{i=1}^na_i.]

3. We have

[ egin{align} sum_{i=1}^{n}(a_i+b_i) &=(a_1+b_1)+(a_2+b_2)+(a_3+b_3)+⋯+(a_n+b_n) [4pt] &=(a_1+a_2+a_3+⋯+a_n)+(b_1+b_2+b_3+⋯+b_n) [4pt] &=sum_{i=1}^na_i+sum_{i=1}^nb_i. end {align}]

□

A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for **sums and powers of integers**, and we use them in the next set of examples.

Rule: Sums and Powers of Integers

1. The sum of (n) integers is given by

[sum_{i=1}^n i=1+2+⋯+n=dfrac{n(n+1)}{2}. label{sum1}]

2. The sum of consecutive integers squared is given by

[sum_{i=1}^n i^2=1^2+2^2+⋯+n^2=dfrac{n(n+1)(2n+1)}{6}. label{sum2}]

3. The sum of consecutive integers cubed is given by

[sum_{i=1}^n i^3=1^3+2^3+⋯+n^3=dfrac{n^2(n+1)^2}{4}. label{sum3} ]

Example (PageIndex{2}): Evaluation Using Sigma Notation

Write using sigma notation and evaluate:

- The sum of the terms ((i−3)^2) for (i=1,2,…,200.)
- The sum of the terms ((i^3−i^2)) for (i=1,2,3,4,5,6)

**Solution**

a. Multiplying out ((i−3)^2), we can break the expression into three terms.

[egin{align*} sum_{i=1}^{200}(i−3)^2 &=sum_{i=1}^{200}(i^2−6i+9) [4pt]

&=sum_{i=1}^{200}i^2−sum_{i=1}^{200}6i+sum_{i=1}^{200}9 [4pt]

&=sum_{i=1}^{200}i^2−6sum_{i=1}^{200}i+sum_{i=1}^{200}9 [4pt]

&=dfrac{200(200+1)(400+1)}{6}−6 left[dfrac{200(200+1)}{2}
ight]+9(200) [4pt]

&=2,686,700−120,600+1800 [4pt]

&=2,567,900 end{align*}]

b. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.

[egin{align*} sum_{i=1}^{6}(i^3−i^2) &=sum_{i=1}^6 i^3−sum_{i=1}^6 i^2 [4pt]

&=dfrac{6^2(6+1)^2}{4}−dfrac{6(6+1)(2(6)+1)}{6} [4pt]

&=dfrac{1764}{4}−dfrac{546}{6} [4pt]

&=350 end{align*} ]

Exercise (PageIndex{2})

Find the sum of the values of (4+3i) for (i=1,2,…,100.)

**Hint**Use the properties of sigma notation to solve the problem.

**Answer**(15,550)

Example (PageIndex{3}): Finding the Sum of the Function Values

Find the sum of the values of (f(x)=x^3) over the integers (1,2,3,…,10.)

**Solution**

Using Equation ef{sum3}, we have

[sum_{i=0}^{10}i^3=dfrac{(10)^2(10+1)^2}{4}=dfrac{100(121)}{4}=3025 onumber]

Exercise (PageIndex{3})

Evaluate the sum indicated by the notation (displaystyle sum_{k=1}^{20}(2k+1)).

- Hint
Use the rule on sum and powers of integers (Equations ef{sum1}- ef{sum3}).

**Answer**(440)

## Approximating Area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let (f(x)) be a continuous, nonnegative function defined on the closed interval ([a,b]). We want to approximate the area (A) bounded by (f(x)) above, the (x)-axis below, the line (x=a) on the left, and the line (x=b) on the right (Figure (PageIndex{1})).

How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval ([a,b]) into (n) subintervals of equal width, (dfrac{b−a}{n}). We do this by selecting equally spaced points (x_0,x_1,x_2,…,x_n) with (x_0=a,x_n=b,) and

[x_i−x_{i−1}=dfrac{b−a}{n}]

for (i=1,2,3,…,n.)

We denote the width of each subinterval with the notation (Δx,) so (Δx=frac{b−a}{n}) and

[x_i=x_0+iΔx]

for (i=1,2,3,…,n.) This notion of dividing an interval ([a,b]) into subintervals by selecting points from within the interval is used quite often in approximating the area under a curve, so let’s define some relevant terminology.

Definition: Partitions

A set of points (P={x_i}) for (i=0,1,2,…,n) with (a=x_0**regular partition **(or uniform partition) of the interval ([a,b].)

We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: the left-endpoint approximation and the right-endpoint approximation.

Rule: Left-Endpoint Approximation

On each subinterval ([x_{i−1},x_i]) (for (i=1,2,3,…,n)), construct a rectangle with width (Δx) and height equal to (f(x_{i−1})), which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is (f(x_{i−1})Δx). Adding the areas of all these rectangles, we get an approximate value for (A) (Figure (PageIndex{2})). We use the notation (L_n) to denote that this is a** left-endpoint approximation** of (A) using (n) subintervals.

[A≈L_n=f(x_0)Δx+f(x_1)Δx+⋯+f(xn−1)Δx=sum_{i=1}^nf(x_{i−1})Δx]

The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

Rule: Right-Endpoint Approximation

Construct a rectangle on each subinterval ([x_{i−1},x_i]), only this time the height of the rectangle is determined by the function value (f(x_i)) at the right endpoint of the subinterval. Then, the area of each rectangle is (f(x_i),Δx) and the approximation for (A) is given by

[A≈R_n=f(x_1)Δx+f(x_2)Δx+⋯+f(x_n)Δx=sum_{i=1}^nf(x_i)Δx.]

The notation (R_n) indicates this is a **right-endpoint approximation** for (A) (Figure (PageIndex{3})).

The graphs in Figure (PageIndex{4}) represent the curve (f(x)=dfrac{x^2}{2}). In Figure (PageIndex{4b}) we divide the region represented by the interval ([0,3]) into six subintervals, each of width (0.5). Thus, (Δx=0.5). We then form six rectangles by drawing vertical lines perpendicular to (x_{i−1}), the left endpoint of each subinterval. We determine the height of each rectangle by calculating (f(x_{i−1})) for (i=1,2,3,4,5,6.) The intervals are ([0,0.5],[0.5,1],[1,1.5],[1.5,2],[2,2.5],[2.5,3]). We find the area of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area between (f(x)) and the (x)-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,

[egin{align*} A≈L_6 &=sum_{i=1}^6f(x_{i−1})Δx =f(x_0)Δx+f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx+f(x_5)Δx [4pt]

&=f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5 [4pt]

&=(0)0.5+(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5 [4pt]

&=0+0.0625+0.25+0.5625+1+1.5625 [4pt]

&=3.4375 , ext{units}^2end{align*} ]

In Figure (PageIndex{4b}), we draw vertical lines perpendicular to (x_i) such that (x_i) is the right endpoint of each subinterval, and calculate (f(x_i)) for (i=1,2,3,4,5,6). We multiply each (f(x_i)) by (Δx) to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under (f(x)). Thus,

[ egin{align*} A≈R_6 &=sum_{i=1}^6f(x_i)Δx=f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx+f(x_5)Δx+f(x_6)Δx[4pt]

&=f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5+f(3)0.5 [4pt]

&=(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5+(4.5)0.5 [4pt]

&=0.0625+0.25+0.5625+1+1.5625+2.25 [4pt]

&=5.6875 , ext{units}^2.end{align*} ]

Example (PageIndex{4}): Approximating the Area Under a Curve

Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of (f(x)=x^2) on the interval ([0,2]); use (n=4).

**Solution**

First, divide the interval ([0,2]) into (n) equal subintervals. Using (n=4,, Δx=dfrac{(2−0)}{4}=0.5). This is the width of each rectangle. The intervals ([0,0.5],[0.5,1],[1,1.5],[1.5,2]) are shown in Figure (PageIndex{5}). Using a left-endpoint approximation, the heights are (f(0)=0,,f(0.5)=0.25,,f(1)=1,) and (f(1.5)=2.25.) Then,

[ egin{align*} L_4 &=f(x_0)Δx+f(x_1)Δx+f(x_2)Δx+f(x_3)Δx [4pt] &=0(0.5)+0.25(0.5)+1(0.5)+2.25(0.5) [4pt] &=1.75 , ext{units}^2 end{align*} ]

The right-endpoint approximation is shown in Figure (PageIndex{6}). The intervals are the same, (Δx=0.5,) but now use the right endpoint to calculate the height of the rectangles. We have

[ egin{align*} R_4 &=f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx [4pt] &=0.25(0.5)+1(0.5)+2.25(0.5)+4(0.5) [4pt] &=3.75 , ext{units}^2 end{align*} ]

The left-endpoint approximation is (1.75, ext{units}^2); the right-endpoint approximation is (3.75 , ext{units}^2).

Exercise (PageIndex{4})

Sketch left-endpoint and right-endpoint approximations for (f(x)=dfrac{1}{x}) on ([1,2]); use (n=4). Approximate the area using both methods.

**Hint**Follow the solving strategy in Example (PageIndex{4}) step-by-step.

**Answer**The left-endpoint approximation is (0.7595 , ext{units}^2). The right-endpoint approximation is (0.6345 , ext{units}^2). See the below Media.

Looking at Figure (PageIndex{4}) and the graphs in Example (PageIndex{4}), we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of (A) will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely.

We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of increasing (n), first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally (32) rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. Figure (PageIndex{7}) shows the area of the region under the curve (f(x)=(x−1)^3+4) on the interval ([0,2]) using a left-endpoint approximation where (n=4.) The width of each rectangle is

[Δx=dfrac{2−0}{4}=dfrac{1}{2}. onumber]

The area is approximated by the summed areas of the rectangles, or

[L_4=f(0)(0.5)+f(0.5)(0.5)+f(1)(0.5)+f(1.5)0.5=7.5 , ext{units}^2 onumber]

Figure (PageIndex{8}) shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure (PageIndex{7}) with this graph with eight rectangles, we can see there appears to be less white space under the curve when (n=8.) This white space is area under the curve we are unable to include using our approximation. The area of the rectangles is

[L_8=f(0)(0.25)+f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)=7.75 , ext{units}^2 onumber]

The graph in Figure (PageIndex{9}) shows the same function with (32) rectangles inscribed under the curve. There appears to be little white space left. The area occupied by the rectangles is

[L_{32}=f(0)(0.0625)+f(0.0625)(0.0625)+f(0.125)(0.0625)+⋯+f(1.9375)(0.0625)=7.9375 , ext{units}^2. onumber]

We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles (Figure (PageIndex{10})), yields an area

[R_4=f(0.5)(0.5)+f(1)(0.5)+f(1.5)(0.5)+f(2)(0.5)=8.5 , ext{units}^2. onumber]

Dividing the region over the interval ([0,2]) into eight rectangles results in (Δx=dfrac{2−0}{8}=0.25.) The graph is shown in Figure (PageIndex{11}). The area is

[R_8=f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)+f(2)(0.25)=8.25 , ext{units}^2 onumber]

Last, the right-endpoint approximation with (n=32) is close to the actual area (Figure (PageIndex{12})). The area is approximately

[R_{32}=f(0.0625)(0.0625)+f(0.125)(0.0625)+f(0.1875)(0.0625)+⋯+f(2)(0.0625)=8.0625 , ext{units}^2 onumber]

Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as (n) gets larger. Furthermore, as (n) increases, both the left-endpoint and right-endpoint approximations appear to approach an area of (8) square units. Table (PageIndex{15}) shows a numerical comparison of the left- and right-endpoint methods. The idea that the approximations of the area under the curve get better and better as (n) gets larger and larger is very important, and we now explore this idea in more detail.

Value of (n) | Approximate Area (L_n) | Approximate Area (R_n) |
---|---|---|

(n=4) | (7.5) | (8.5) |

(n=8) | (7.75) | (8.25) |

(n=32) | (7.94) | (8.06) |

### Forming Riemann Sums

So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval ([x_{i−1},x_i]). In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point (x^∗_i) in the subinterval ([x_{i−1},x_i]), and use (f(x^∗_i)) as the height of our rectangle. This gives us an estimate for the area of the form

[A≈sum_{i=1}^nf(x^∗_i),Δx.]

A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.

Definition: Riemann sum

Let (f(x)) be defined on a closed interval ([a,b]) and let (P) be any partition of ([a,b]). Let (Δx_i) be the width of each subinterval ([x_{i−1},x_i]) and for each (i)*,* let (x^∗_i) be any point in ([x_{i−1},,x_i]). A Riemann sum is defined for (f(x)) as

[sum_{i=1}^nf(x^∗_i),Δx_i.]

At this point, we'll choose a regular partition (P), as we have in our examples above. This forces all (Δx_i) to be equal to (Δx = dfrac{b-a}{n}) for any natural number of intervals (n).

Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as (n) get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of (n). We are now ready to define the area under a curve in terms of Riemann sums.

Definition: Area Under the Curve

Let (f(x)) be a continuous, nonnegative function on an interval ([a,b]), and let (displaystyle sum_{i=1}^nf(x^∗_i),Δx) be a Riemann sum for (f(x)) with a regular partition (P). Then, the **area under the curve** (y=f(x)) on ([a,b]) is given by

[A=lim_{n→∞}sum_{i=1}^nf(x^∗_i),Δx.]

See a graphical demonstration of the construction of a Riemann sum.

Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function (f(x)) as (x) goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.

Second, we must consider what to do if the expression converges to different limits for different choices of ({x^∗_i}.) Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if (f(x)) is continuous on the closed interval ([a,b]), then (displaystyle lim_{n→∞}sum_{i=1}^nf(x^∗_i)Δx) exists and is unique (in other words, it does not depend on the choice of ({x^∗_i})).

We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for ({x^∗_i}). Although any choice for ({x^∗_i}) gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for ({x^∗_i}) to guarantee one result or the other.

If we want an overestimate, for example, we can choose ({x^∗_i}) such that for (i=1,2,3,…,n,) (f(x^∗_i)≥f(x)) for all (x∈[x_i−1,x_i]). In other words, we choose ({x^∗_i}) so that for (i=1,2,3,…,n,) (f(x^∗_i)) is the maximum function value on the interval ([x_{i−1},x_i]). If we select ({x^∗_i}) in this way, then the Riemann sum (displaystyle sum_{i=1}^nf(x^∗_i)Δx) is called an **upper sum**. Similarly, if we want an underestimate, we can choose ({x∗i}) so that for (i=1,2,3,…,n,) (f(x^∗_i)) is the minimum function value on the interval ([x_{i−1},x_i]). In this case, the associated Riemann sum is called a **lower sum**. Note that if (f(x)) is either increasing or decreasing throughout the interval ([a,b]), then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.

Example (PageIndex{5}): Finding Lower and Upper Sums

Find a lower sum for (f(x)=10−x^2) on ([1,2]); let (n=4) subintervals.

**Solution**

With (n=4) over the interval ([1,2], ,Δx=dfrac{1}{4}). We can list the intervals as ([1,1.25],,[1.25,1.5],,[1.5,1.75],) and ([1.75,2]). Because the function is decreasing over the interval ([1,2],) Figure shows that a lower sum is obtained by using the right endpoints.

The Riemann sum is

[egin{align*} sum_{k=1}^4(10−x^2)(0.25) &=0.25[10−(1.25)^2+10−(1.5)^2+10−(1.75)^2+10−(2)^2] [4pt]

&=0.25[8.4375+7.75+6.9375+6] [4pt]

&=7.28 , ext{units}^2.end{align*}]

The area of (7.28) ( ext{units}^2) is a lower sum and an underestimate.

Exercise (PageIndex{5})

- Find an upper sum for (f(x)=10−x^2) on ([1,2]); let (n=4.)
- Sketch the approximation.

**Hint**(f(x)) is decreasing on ([1,2]), so the maximum function values occur at the left endpoints of the subintervals.

**Answer**a. Upper sum=(8.0313 , ext{units}^2.)

b.

Example (PageIndex{6}): Finding Lower and Upper Sums for (f(x)=sin x)

Find a lower sum for (f(x)=sin x) over the interval ([a,b]=left[0,frac{π}{2} ight]); let (n=6.)

**Solution**

Let’s first look at the graph in Figure (PageIndex{14}) to get a better idea of the area of interest.

The intervals are (left[0,frac{π}{12} ight],,left[frac{π}{12},frac{π}{6} ight],,left[frac{π}{6},frac{π}{4} ight],,left[frac{π}{4},frac{π}{3} ight],,left[frac{π}{3},frac{5π}{12} ight]), and (left[frac{5π}{12},frac{π}{2} ight]). Note that (f(x)=sin x) is increasing on the interval (left[0,frac{π}{2} ight]), so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum (sum_{i=0}^5sin x_ileft( frac{π}{12} ight)).We have

[A≈sin(0)left( frac{π}{12} ight)+sinleft( frac{π}{12} ight)left( frac{π}{12} ight)+sinleft( frac{π}{6} ight)left( frac{π}{12} ight)+sinleft( frac{π}{4} ight)left( frac{π}{12} ight)+sinleft( frac{π}{3} ight)left( frac{π}{12} ight)+sinleft( frac{5π}{12} ight)left( frac{π}{12} ight)approx 0.863 , ext{units}^2. onumber]

Exercise (PageIndex{6})

Using the function (f(x)=sin x) over the interval (left[0,frac{π}{2} ight],) find an upper sum; let (n=6.)

**Hint**Follow the steps from Example (PageIndex{6}).

**Answer**(A≈1.125 , ext{units}^2)

## Key Concepts

- The use of sigma (summation) notation of the form (displaystyle sum_{i=1}^na_i) is useful for expressing long sums of values in compact form.
- For a continuous function defined over an interval ([a,b],) the process of dividing the interval into (n) equal parts, extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing the areas yields an approximation of the area of that region.
- When using a regular partition, the width of each rectangle is (Δx=dfrac{b−a}{n}).
- Riemann sums are expressions of the form (displaystyle sum_{i=1}^nf(x^∗_i)Δx,) and can be used to estimate the area under the curve (y=f(x).) Left- and right-endpoint approximations are special kinds of Riemann sums where the values of ({x^∗_i}) are chosen to be the left or right endpoints of the subintervals, respectively.
- Riemann sums allow for much flexibility in choosing the set of points ({x^∗_i}) at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum.

## Key Equations

**Properties of Sigma Notation**

[egin{align*} sum_{i=1}^nc&=nc [4pt]

sum_{i=1}^nca_i &=csum_{i=1}^na_i [4pt]

sum_{i=1}^n(a_i+b_i) &=sum_{i=1}^na_i+sum_{i=1}^nb_i [4pt]

sum_{i=1}^n(a_i−b_i) &=sum_{i=1}^na_i−sum_{i=1}^nb_i [4pt]

sum_{i=1}^na_i&=sum_{i=1}^ma_i+sum_{i=m+1}^na_i end{align*}]

**Sums and Powers of Integers**

[sum_{i=1}^ni=1+2+⋯+n=dfrac{n(n+1)}{2} onumber ]

[sum_{i=1}^ni^2=1^2+2^2+⋯+n^2=dfrac{n(n+1)(2n+1)}{6} onumber]

[sum_{i=0}^ni^3=1^3+2^3+⋯+n^3=dfrac{n^2(n+1)^2}{4} onumber]

**Left-Endpoint Approximation**

(A≈L_n=f(x_0)Δx+f(x_1)Δx+⋯+f(x_{n−1})Δx=displaystyle sum_{i=1}^nf(x_{i−1})Δx)

**Right-Endpoint Approximation**

(A≈R_n=f(x_1)Δx+f(x_2)Δx+⋯+f(x_n)Δx=displaystyle sum_{i=1}^nf(x_i)Δx)

## Glossary

**left-endpoint approximation**- an approximation of the area under a curve computed by using the left endpoint of each subinterval to calculate the height of the vertical sides of each rectangle

**lower sum**- a sum obtained by using the minimum value of (f(x)) on each subinterval

**partition**- a set of points that divides an interval into subintervals

**regular partition**- a partition in which the subintervals all have the same width

**riemann sum**- an estimate of the area under the curve of the form (A≈displaystyle sum_{i=1}^nf(x^∗_i)Δx)

**right-endpoint approximation**- the right-endpoint approximation is an approximation of the area of the rectangles under a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle

**sigma notation**- (also,
**summation notation**) the Greek letter sigma ((Σ)) indicates addition of the values; the values of the index above and below the sigma indicate where to begin the summation and where to end it

**upper sum**- a sum obtained by using the maximum value of (f(x)) on each subinterval

## Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

## 5.5 Land Surveying and Conventional Techniques for Measuring Positions on the Earth’s Surface

Ease, accuracy, and worldwide availability have made ‘GPS’ a household term. Yet, none of the power or capabilities of GPS would have been possible without traditional surveyors paving the way. The techniques and tools of conventional surveying are still in use and, as you will see, are based on the very same concepts that underpin even the most advanced satellite-based positioning.

Geographic positions are specified relative to a fixed reference. Positions on the globe, for instance, may be specified in terms of angles relative to the center of the Earth, the equator, and the prime meridian.

Land surveyors measure horizontal positions in geographic or plane coordinate systems relative to previously surveyed positions called **control points,** most of which are indicated physically in the world with a metal “benchmark” that fixes the location and, as shown here, may also indicate elevation about mean sea level (Figure 5.10). In 1988 NGS established **four orders of control point accuracy**, ranging in maximum base error from 3mm to 5cm. In the U.S., the National Geodetic Survey (NGS) maintains a **National Spatial Reference System (NSRS)** that consists of approximately 300,000 horizontal and 600,000 vertical control stations (Doyle,1994).

Doyle (1994) points out that horizontal and vertical reference systems coincide by less than ten percent. This is because:

. horizontal stations were often located on high mountains or hilltops to decrease the need to construct observation towers usually required to provide line-of-sight for triangulation, traverse and trilateration measurements. Vertical control points however, were established by the technique of spirit leveling which is more suited to being conducted along gradual slopes such as roads and railways that seldom scale mountain tops. (Doyle, 2002, p. 1)

You might wonder how a control network gets started. If positions are measured relative to other positions, what is the first position measured relative to? The answer is: the stars. Before reliable timepieces were available, astronomers were able to determine longitude only by careful observation of recurring celestial events, such as eclipses of the moons of Jupiter. Nowadays, geodesists produce extremely precise positional data by analyzing radio waves emitted by distant stars. Once a control network is established, however, **surveyors produce positions using instruments that measure angles and distances between locations on the Earth's surface.**

### 5.5.1 Measuring Angles and Distances

You probably have seen surveyors working outside, e.g., when highways are being realigned or new housing developments are being constructed. Often one surveyor operates equipment on a tripod while another holds up a rod some distance away. What the surveyors and their equipment are doing is carefully measuring angles and distances, from which positions and elevation can be calculated. We will briefly discuss this equipment and their methodology. Let us first take a look at angles and how they apply to surveying.

Although a standard compass can give you a rough estimate of angles, the Earth’s magnetic field is not constant and the magnetic poles, which slowly move over time, do not perfectly align with the planet’s axis of rotation as a result of the latter, true (geographic) north and magnetic north are different. Moreover, some rocks can become magnetized and introduce subtle local anomalies when using compass. For these reasons, land surveyors rely on **transits** (or their more modern equivalents, called **theodolites**) to measure angles. A transit (Figure 5.11) consists of a telescope for sighting distant target objects, two measurement wheels that work like protractors for reading horizontal and vertical angles, and bubble levels to ensure that the angles are true. A theodolite is essentially the same instrument, except that it is somewhat more complex and capable of higher precision. In modern theodolites, some mechanical parts are replaced with electronics.

When surveyors measure angles, the resultant calculations are typically reported as either **azimuths** or **bearings,** as seen in Figure 5.12. A bearing is an angle less than 90° within a quadrant defined by the cardinal directions. An azimuth is an angle between 0° and 360° measured clockwise from North. "South 45° East" and "135°" are the same direction expressed as a bearing and as an azimuth.

### 5.5.2 Measuring Distances

To measure distances, land surveyors once used 100-foot long metal tapes that are graduated in hundredths of a foot. An example of this technique is shown in Figure 5.13. Distances along slopes were measured in short horizontal segments. Skilled surveyors could achieve accuracies of up to one part in 10,000 (1 centimeter error for every 100 meters distance). Sources of error included flaws in the tape itself, such as kinks variations in tape length due to extremes in temperature and human errors such as inconsistent pull, allowing the tape to stray from the horizontal plane, and incorrect readings.

Since the 1980s, **electronic distance measurement(EDM)** devices have allowed surveyors to measure distances more accurately and more efficiently than they can with tapes. To measure the horizontal distance between two points, one surveyor uses an EDM instrument to shoot an energy wave toward a reflector held by the second surveyor. The EDM records the elapsed time between the wave's emission and its return from the reflector. It then calculates distance as a function of the elapsed time (not unlike what we’ve learned about GPS!). Typical short-range EDMs can be used to measure distances as great as 5 kilometers at accuracies up to one part in 20,000, twice as accurate as taping.

Instruments called **total stations** (Figure 5.14) combine electronic distance measurement and the angle measuring capabilities of theodolites in one unit. Next we consider how these instruments are used to measure horizontal positions in relation to established control networks.

### 5.5.3 Combining Angles and Distances to Determine Positions

Surveyors have developed distinct methods, based on separate control networks, for measuring horizontal and vertical positions. In this context, a horizontal position is the location of a point relative to two axes: the equator and the prime meridian on the globe, or to the x and y axes in a plane coordinate system.

We will now introduce two techniques that surveyors use to create and extend control networks (triangulation and trilateration) and two other techniques used to measure positions relative to control points (open and closed traverses).

Surveyors typically measure positions in series. Starting at control points, they measure angles and distances to new locations, and use trigonometry to calculate positions in a plane coordinate system. Measuring a series of positions in this way is known as "running a traverse." A traverse that begins and ends at different locations, in which at least one end point is initially unknown, is called an open traverse. A traverse that begins and ends at the same point, or at two different but known points, is called a closed traverse. "Closed" here does not mean geometrically closed (as in a polygon) but mathematically closed (defined as: of or relating to an interval containing both its endpoints). By "closing" a route between one known location and another known location, the surveyor can determine errors in the traverse.

Measurement errors in a closed traverse that connects at the point where it started can be quantified by summing the interior angles of the polygon formed by the traverse. The accuracy of a single angle measurement cannot be known, but since the sum of the interior angles of a polygon is always (n-2) × 180, it's possible to evaluate the traverse as a whole, and to distribute the accumulated errors among all the interior angles. Errors produced in an open traverse, one that does not end where it started, cannot be assessed or corrected. The only way to assess the accuracy of an open traverse is to measure distances and angles repeatedly, forward and backward, and to average the results of calculations. Because repeated measurements are costly, other surveying techniques that enable surveyors to calculate and account for measurement error are preferred over open traverses for most applications.

### 5.5.4 Triangulation

Closed traverses yield adequate accuracy for property boundary surveys, provided that an established control point is nearby. Surveyors conduct control surveys to extend and add point density to horizontal control networks. Before survey-grade satellite positioning was available, the most common technique for conducting control surveys was **triangulation** (Figure 5.16).

- Using a total station equipped with an electronic distance measurement device, the control survey team commences by measuring the azimuth
*alpha*, and the baseline distance AB. - These two measurements enable the survey team to calculate position B as in an open traverse.
- The surveyors next measure the interior angles CAB, ABC, and BCA at point A, B, and C. Knowing the interior angles and the baseline length, the trigonometric "law of sines" can then be used to calculate the lengths of any other side. Knowing these dimensions, surveyors can fix the position of point C.
- Having measured three interior angles and the length of one side of triangle ABC, the control survey team can calculate the length of side BC. This calculated length then serves as a baseline for triangle BDC. Triangulation is thus used to extend control networks, point by point and triangle by triangle.

### 5.5.5 Trilateration

An alternative to triangulation is **trilateration**, which uses distances alone to determine positions. By eschewing angle measurements, trilateration is easier to perform, requires fewer tools, and is therefore less expensive. Having read this chapter so far, you have already been introduced to a practical application of trilateration, since it is the technique behind satellite ranging used in GPS.

You have seen an example of trilateration in Figure 5.8 in the form of 3-dimensional spheres extending from orbiting satellites. Demo 1 below steps through this process in two dimensions.

#### Try This: Step through the process of 2-dimensional trilateration.

Once a distance from a control point is established, a person can calculate a distance by open traverse, or rely on a known distance if one exists. A single control point and known distance confines the possible locations of an unknown point to the edge of the circle surrounding the control point at that distance there are infinitively many possibilities along this circle for the unknown location. The addition of a second control point introduces another circle with a radius equal to its distance from the unknown point. With two control points and distance circles, the number of possible points for the unknown location is reduced to exactly two. A third and final control point can be used to identify which of the remaining possibilities is the true location.

Trilateration is noticeably simpler than triangulation and is a very valuable skill to possess. Even with very rough estimates, one can determine a general location with reasonable success.

#### Practice Quiz

Registered Penn State students should return now take the self-assessment quiz **Land Surveying**.

You may take practice quizzes as many times as you wish. They are not scored and do not affect your grade in any way.

## Audio Level Adjustments

Level calibration between all of your speakers in a 5.1 system is one of the most important adjustable parameters. Many people tend to boost the rear speaker levels too high relative to the fronts and center channels. This sometimes tends to overemphasis the rear channels resulting in an unnatural surround field that is easily localized by the ear. Over emphasizing a particular channel sound level will diminish the balance in the system. Doing so undermines what the recording engineering intended the mix to sound like for the movie and/or music CD.

### I recommend the following procedure for proper level calibration of your 5.1 Surround System:

This is the first step in calibrating the sound levels of your 5.1 system. Please remember that these levels will need to be tweaked depending on the source recording and/or surround scheme.

**For example:** Some recordings may boost the rear channels 1-3db higher than nominal due to poor mixing methods or deliberate wow factor. If the rear levels sound too loud when watching a movie or listening to a 5.1/6.1/7.1 audio mix, simply lower them until they sound balanced relative to the front and center channels.

I recommend acquiring a multi-channel set-up disc such as the one from Avia, Sound & Vision, or DTS. It will help you calibrate your listening levels of your speakers and subwoofer as it sweeps frequencies 20Hz to 20KHz for all channels.

If you don't trust your ears, you may wish to purchase a Radio Shack SPL meter. When operating the test tone of your Receiver/Preamp, calibrate the volume levels within 1 dB relative to each channel. Hold the unit so that the microphone is pointed at the ceiling and position the microphone as closely to ear level as possible (at the sweet spot listening position) when running this test.

**Note:** Make sure you set the scale to "C-Weighted" on the SPL meter as this closely matches a flat frequency response curve throughout the audible band. The early AVIA test discs used to recommend the "fast" setting on the SPL meter but we've found it easier to get more consistent results using the "slow' setting which allows easier readability.

**Technical Advances Update:** Most receivers/processors built since 2003-2004 enable you to merely configure the *distance* from each speaker to the listening position. The mathematical delay calculations are then performed by the receiver or pre-processor, thus saving you a lot of headache and time. be sure to calculate distances to a singular point, the sweet spot, even though you may have listening positions throughout the listening room.

##### About the author:

Gene manages this organization, establishes relations with manufacturers and keeps Audioholics a well oiled machine. His goal is to educate about home theater and develop more standards in the industry to eliminate consumer confusion clouded by industry snake oil.

Confused about what AV Gear to buy or how to set it up? Join our Exclusive Audioholics E-Book Membership Program!

## Using snapping

The Measure tool uses snapping in ArcMap—the cursor will snap to features, edges, and coordinates that you have specified in your snapping settings. When you place your pointer on your map document and begin to enter coordinates, your snapping settings will be used. To learn more about snapping, see About snapping.

When you use the Measure tool in snapping mode, it is easy to trace over features, such as to measure the distance between street intersections. If you want to snap to edges (the parts of lines where there is no vertex, such as the edge of a parcel with straight sides), hold down CTRL when measuring. To turn of snapping temporarily, hold down the SPACEBAR key.

## Parameters

The point features from which distances to the near features will be calculated.

The points to which distances from the input features will be calculated. Distances between points within the same feature class or layer can be determined by specifying the same feature class or layer for the input and near features.

The table containing the list of input features and information about all near features within the search radius. If a search radius is not specified, distances from all input features to all near features are calculated.

Specifies the radius used to search for candidate near features. The near features within this radius are considered for calculating the nearest feature. If no value is specified (that is, the default (empty) radius is used) all near features are considered for calculation. The unit of search radius defaults to units of the input features. The units can be changed to any other unit. However, this has no impact on the units of the output DISTANCE field which is based on the units of the coordinate system of the input features.

The point features from which distances to the near features will be calculated.

The points to which distances from the input features will be calculated. Distances between points within the same feature class or layer can be determined by specifying the same feature class or layer for the input and near features.

The table containing the list of input features and information about all near features within the search radius. If a search radius is not specified, distances from all input features to all near features are calculated.

Specifies the radius used to search for candidate near features. The near features within this radius are considered for calculating the nearest feature. If no value is specified (that is, the default (empty) radius is used) all near features are considered for calculation. The unit of search radius defaults to units of the input features. The units can be changed to any other unit. However, this has no impact on the units of the output DISTANCE field which is based on the units of the coordinate system of the input features.

### Code sample

The following Python interactive window script demonstrates how to use the PointDistance function in immediate mode.

The following Python script demonstrates how to use the PointDistance function in a stand-alone script.

## Height of Hazard

The first variable to consider is the height of the hazard. If the hazard is sufficiently far above the ground or expected working surface then guarding is not required. The relevant OSHA standard, 29 CFR 1910.219 – Mechanical Power-Transmission Apparatus, says that a hazard that is more than 7 feet from the working surface does not need to be guarded.

Figure 1 - Height of Hazard Zone |

American National Standards Institute (ANSI) B11.19-2010 – Performance Criteria for Safeguarding requires that a low-risk hazard needs to be safeguarded unless it is 2,500 mm (98.4 in.) or more from the working surface and that a high-risk hazard needs to be safeguarded unless it is 2,700 mm (106.3 in.) or more from the reference plane as shown in Figure 1. This portion of ANSI B11.19-2010 has been harmonized with Canadian Standards Association (CSA) Z432-04 – Safeguarding of Machinery – Occupational Health and Safety, and ISO 13857:2008 – Safety of Machinery.

### Barrier Guards

Barrier (hard) guards can provide the maximum amount of protection by keeping people out, and protecting people outside the hazardous area from objects flying from the machine. Barrier guards normally are not used for the entire perimeter, as this would make it very difficult to access the equipment. Typically, there is an entry to the cell and a more flexible guarding solution will enable personnel or material to safely approach the equipment.

One approach is the use of movable barrier guards in conjunction with devices interlocked with the machine controls to control the hazard(s) whenever the guard is open. When guard door movement is detected, the interlock device sends a stop signal to the guarded equipment. Interlock switches may incorporate a solenoid device that locks the guard door closed and will not release it until the hazardous machine motion has ceased.

A guard must ensure that individuals cannot reach the hazard by reaching over, under, around or through it. To determine the safe mounting distance for a barrier guard, first consider the largest opening in the guarding material. The current OSHA standard for safe distance as a function of opening size is set forth in Table O-10 of OSHA 29 CFR 1910.217 – Mechanical Power Presses. This table technically only applies to mechanical power presses operated within OSHA’s jurisdiction, although some industry consensus standards also reference this table, such as ANSI B65.1-2005, a safety standard for printing press systems.

Figure 2 - ANSI Standard Distance for Barriers |

A 1995 study, A Review of Machine Guarding Recommendations, conducted by the Liberty Mutual Research Center for Health and Safety, is the basis for the ANSI and CSA standards. This anthropomorphic study, shown in Figure 2, was based on the then-current U.S. work force. While not officially adopted by OSHA, these standards have been adopted by a number of other consensus standards, including (among others):

- ANSI B11.19-2010 – Performance Criteria for Safeguarding
- ANSI/RIA R15.06-1999 (R2009) – For Industrial Robots and Robot Systems – Safety Requirements
- CSA Z432-04 – Safeguarding of Machinery – Occupational Health and Safety.

**Reach under** – To prevent an individual from accessing the hazard by reaching or crawling below the barrier guard, perimeter barrier guards must be designed so that the bottom of the barrier is no more than 300 mm (1 foot) above the adjacent walking surface according to ANSI/RIA R15.06-1999 (R2009). The same standard states that the top of the barrier must be no less than 1,500 mm (5 feet) above the adjacent walking surface. These measurements are more restrictive in Canada with distances of 150 mm (6 inches) and 1,800 mm (71 inches), respectively, according to CSA Z434-03. The equivalent international standard, ISO 10218-2:2011, sets the requirements at 200 mm (7.8 inches) and 1,400 mm (55 inches), respectively.

Figure 3 - Guard Height to Protect Against Reach Over |

**Reach over** – Figure 3 shows how to determine guard height to protect against reaching over a barrier to contact a hazard. In the figure, “a” is the height of the danger zone, “b” is the height of the protective structure and “c” is the horizontal distance between the guard and the danger zone. Guards or other protective structures less than 1,000 mm high (39 inches) are not considered sufficient on their own for any application because they do not adequately restrict movement of the body, and structures less than 1,400 mm high (55 inches) should not be used in high-risk applications without additional safety measures. The following guidelines are available to help determine adequate height of constructed guards in relation to the hazard height and the distance of the guard from hazard:

- ANSI B11.19-2010 – Performance Criteria for Safeguarding
- CSA Z432-04 – Safeguarding of Machinery – Occupational Health and Safety
- ISO 13857:2008 – Safety of Machinery – Safety distances to prevent hazard zones being reached by upper and lower limbs.
- Note that ANSI B15.1-2000 (R2006) – Safety Standard for Mechanical Power Transmission Apparatus included similar requirements, but has since been withdrawn and replaced in part by ANSI B11.19-2010.

## 5.5 Contour Lines and Intervals

A contour line is a line drawn on a topographic map to indicate ground elevation or depression. A contour interval is the vertical distance or difference in elevation between contour lines. Index contours are bold or thicker lines that appear at every fifth contour line.

If the numbers associated with specific contour lines are increasing, the elevation of the terrain is also increasing. If the numbers associated with the contour lines are decreasing, there is a decrease in elevation. As a contour approaches a stream, canyon, or drainage area, the contour lines turn upstream. They then cross the stream and turn back along the opposite bank of the stream forming a "v". A rounded contour indicates a flatter or wider drainage or spur. Contour lines tend to enclose the smallest areas on ridge tops, which are often narrow or very limited in spatial extent. Sharp contour points indicate pointed ridges.

Example 1 - In the graphic below, what is the vertical distance between the contour lines?

Pick two contour lines that are next to each other and find the difference in associated numbers.

40 feet - 20 feet = 20 feet

The contour lines in this figure are equally spaced. The even spacing indicates the hill has a uniform slope. From the contour map, a profile can be drawn of the terrain.

**Example 2** - Draw a profile showing the elevations of the contours.

Note: The intervals are increasing, therefore, the contours indicate a hill. The peak is normally considered to be located at half the interval distance.

Widely separated contour lines indicate a gentle slope. Contour lines that are very close together indicate a steep slope.

The figure above illustrates various topographic features. (b) Notice how a mountain saddle, a ridge, a stream, a steep area, and a flat area are shown with contour lines.

The figure above illustrates a depression and its representation using contour lines. Notice the tick marks pointing toward lower elevation.

Select the correct answer(s) from the questions below:

### SLOPE PERCENT FROM TOPOGRAPHIC MAP

The horizontal distance between points A and B can be measured with a scaled ruler and used to determine the slope percent.

slope percent = rise/run × 100

Example 4 - What is the slope percent in Exercise 2 above?

slope percent = rise/run × 100.

For this computation, the rise, or vertical ground distance, and run, or horizontal ground distance, are needed.

__Step 1.__ Measure the horizontal map distance between points A and B to get the vertical ground distance.

The horizontal map distance measures 0.5 inches.

__Step 2.__ Use the appropriate conversion factor to convert the horizontal map distance to horizontal ground distance.

0.5 in × 24,000 in/in = 12,000 in

__Step 3.__ The desired unit is feet. Set up the cancellation table so all units will cancel, except the desired unit, feet.

__Step 4.__ Use the slope percent equation and solve. The run is 1000 feet and the rise in elevation is 120 feet.

slope percent = rise/run × 100

slope percent = (120ft / 1000ft) × 100 = 12%

Slope Worksheet - Use the information from the example above and complete the slope worksheet. Line 1 starts with the contour interval, not the projection point.

## Measure distances and areas in Google Earth

You can measure distances between locations and along paths. You can also measure the size of polygons that you draw in Google Earth.

- Measurements may not be 100% accurate, especially in areas with 3D terrain and buildings. For best results, measure using a top-down view.
- Measurements don’t account for changes in elevation.
- These instructions only apply to the new Google Earth. Learn how to measure distances in Google Earth Pro.

- On your Android phone or tablet, open the Google Earth app .
- Search for a place, or select a location on the globe.
- Tap Measure .
- To add measurement points, move the map and tap
**Add point.** - To remove a point, at the top, tap Undo .
- When finished, at the top, tap Done . On the bottom, you’ll see the distance measurement.

**Note**: If you also want to measure the area of a location, connect to your first point and tap **Close Shape**.

## Hands-on Activity Using Map Scales to Figure Distances and Areas

Units serve as guides to a particular content or subject area. Nested under units are lessons (in purple) and hands-on activities (in blue).

Note that not all lessons and activities will exist under a unit, and instead may exist as "standalone" curriculum.

### TE Newsletter

Students use map scales to determine map distances and areas,

### Summary

### Engineering Connection

Many types of engineers—civil, geological, petroleum, environmental—must fully understand maps, map reading and map creation to assist in the research and planning of engineering design solutions, such as designing roadways and tunnels, drilling for water or fossil fuels, creating dams, and tracking air pollution.

### Learning Objectives

After this activity, students should be able to:

- Use a map scale to determine distances between cities on a map and the size of areas on a map.
- Compare map areas determined in this activity to areas from a previous activity.
- Relate map areas and lengths to a real-world situation.

### Educational Standards

Each *TeachEngineering* lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards.

All 100,000+ K-12 STEM standards covered in *TeachEngineering* are collected, maintained and packaged by the *Achievement Standards Network (ASN)*, a project of *D2L* (www.achievementstandards.org).

In the ASN, standards are hierarchically structured: first by source *e.g.*, by state within source by type *e.g.*, science or mathematics within type by subtype, then by grade, *etc*.

###### NGSS: Next Generation Science Standards - Science

MS-ESS2-2. Construct an explanation based on evidence for how geoscience processes have changed Earth's surface at varying time and spatial scales. (Grades 6 - 8)

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Water's movements—both on the land and underground—cause weathering and erosion, which change the land's surface features and create underground formations.

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###### Common Core State Standards - Math

- Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation. (Grade 6) More Details

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###### International Technology and Engineering Educators Association - Technology

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###### State Standards

###### Colorado - Math

- Fluently add, subtract, multiply, and divide multidigit decimals using standard algorithms for each operation. (Grade 6) More Details

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###### Colorado - Science

- Develop and communicate an evidence based scientific explanation around one or more factors that change Earth's surface (Grade 5) More Details

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### Materials List

- rulers, meter sticks, measuring tapes
- calculators
- 4 handouts: Scaling the Map Worksheet, Alabraska General Map, Alabraska Geology Map, Reference Page (vocabulary, formulas, unit conversions)
- (optional) a large-sized Alabraska General Map, 2 x 3 ft (.6 x 1 m), for display in the classroom alternatively, use a projector to enlarge and display the map

### Worksheets and Attachments

### More Curriculum Like This

Students apply their knowledge of scales and areas to determine the best locations in Alabraska for the underground caverns. They cut out rectangular paper pieces to represent caverns to scale with the maps and place the cutouts on the maps to determine possible locations.

Through this earth science curricular unit composed of eight activities, student teams are presented with the scenario that an asteroid will impact the Earth. In response, their challenge is to design the location and size of underground caverns to shelter the people from an uninhabitable Earth for .

To kickoff the Adventure Engineering Asteroid Impact unit, students learn of the impending asteroid impact scenario, form teams and begin to study the situation in depth. A simple in-class simulation shows them the potential for destruction and disaster. They look at maps and complete a worksheet an.

### Pre-Req Knowledge

Some knowledge of length, width, area and volume multiplication.

### Introduction/Motivation

Now that your engineering team knows the required cavern area, your task is to translate that information to a map to see how big the cavern is compared to the size of the state of Alabraska.

### Procedure

As necessary, refresh students' knowledge on how to use map scales.

- Gather materials and make copies of the handouts.
- Post (or project) the large-sized Alabraska General Map at the front of the classroom.

- Hand out the materials to the groups.
- Re-familiarize students with the maps by asking them a few questions, for example: What is the capitol of Alabraska? Where is it located in terms of grid coordinates? What types of transportation are represented in the state of Alabraska?
- Discuss the map scale with students. Lead them through some examples by making map measurements with a ruler and then determining how many miles this represents according to the map scale. For example, if 1 centimeter = 10 kilometers, then 3 cm on the map represents 30 kms in the real world.
- Give the engineering teams time to complete the worksheet.
*Tips*for worksheet questions:

Q1: To put the worksheet answers into perspective, ask students to compare their answers to their own home-to-school distance.

Q2:The grid space area is determined by multiplying length by width.

Q3: Expect students to figure out that they can count grid spaces within the military base and multiply by the area per grid space (Q2 answer).

Q4: This is to give the students perspective.

Q5: Expect students to find that Alabraska is much bigger than the required cavern area.

- As time permits or assign as homework: If students drew plans of their cavern designs at the end of the previous activity, now have them re-draw them to scale. An appropriate scale is 1 cm = 1 km.

### Assessment

*Worksheet*: Review students' answers on the Scaling the Map Worksheet to gauge their mastery of the subject matter. Refer to the *Asteroid Impact Student Workbook Example Answers* provided in the unit document for example worksheet answers.

*Quiz*: To conclude, ask students to estimate distances using a different map. For example, estimate the distance from two cities such as London and Moscow, or the area of a U.S. state. Or have students conduct the Extension Activities.

*Homework*: If students drew plans of their cavern designs at the end of the previous activity, now assign them re-draw them to scale. An appropriate scale is 1 cm = 1 km.

### Activity Extensions

- On a map of the U.S., use the scale to determine the area of any state.
- Find the largest country in the world and use the scale to determine its size.
- What are the tallest buildings in the world? For example, the Willis Tower in Chicago is 1,450 feet tall (110 floors). If you were asked to build a model of this skyscraper at a scale of 1 inch = 100 feet, how tall would the model be? (Answer: 14.5 inches tall.) For extra credit, convert English units to metric units (442 meters).

### Copyright

### Supporting Program

### Acknowledgements

Adventure Engineering was supported by National Science Foundation grant nos. DUE 9950660 and GK-12 0086457. However, these contents do not necessarily represent the policies of the National Science Foundation, and you should not assume endorsement by the federal government.

## Table of Distances

The Federal explosive regulations require explosives storage magazines to be located certain minimum distances from inhabited buildings, public highways, passenger railways, and other magazines based on the quantity of explosive materials in each magazine. These tables of distances were adopted to protect the public in the event of a magazine explosion.

*Tables of distances apply to the outdoor storage of explosive materials.*

*When determining the distance from a magazine to a highway, an individual must measure from the nearest edge of the magazine to the nearest edge of the highway.*

*If any two or more magazines are separated by less than the specified distance, then the weights in the magazines must be combined and considered as one.*

*Each type of explosive has a specific table of distance.*

### Tables of Distances

#### Explosives

#### Fireworks

### Explosives

#### Applying Table of Distances at § 555.218 and § 555.220

The keys to applying these tables to donor/acceptor relationships are the net explosive weight (NEW) of the donor the distances between magazines the type of materials in the donor magazine and the type of materials in the acceptor magazine. **When storing high explosives (HE), blasting agents (BA) and ammonium nitrate (AN):**

Multiply the minimum distance by 6 if unbarricaded.

Use the proper column for acceptor (AN) (reduced sensitivity of AN acceptor is accounted for by table)?AN cannot be the donor in this relationship

Use the proper column for the acceptor (BA or AN) (reduced sensitivity of acceptor is accounted for by table)

Multiply distance by 6 if unbarricaded.

Use the table at 555.218 to determine the required distance for the storage of blasting agents and ammonium nitrate from inhabited buildings, highways and passenger railways.

#### § 555.218 Table of distances for storage of explosive materials (high)

When two or more storage magazines are located on the same property, each magazine must comply with the minimum distances specified from inhabited buildings, railways, and highways, and, in addition, they should be separated from each other by not less than the distances shown for "Separation of Magazines," except that the quantity of explosives contained in cap magazines shall govern in regard to the spacing of said cap magazines from magazines containing other explosives. If any two or more magazines are separated from each other by less than the specified "Separation of Magazines" distances, then such two or more magazines, as a group, must be considered as one magazine.

#### § 555.219 Table of distances for storage of low explosives

Pounds Over | Pounds Not Over | From Inhabited building distance (feet) | From public railroad and highway distance (feet) | From above ground magazine (feet) |
---|---|---|---|---|

0 | 1,000 | 75 | 75 | 50 |

1,000 | 5,000 | 115 | 115 | 75 |

5,000 | 10,000 | 150 | 150 | 100 |

10,000 | 20,000 | 190 | 190 | 125 |

20,000 | 30,000 | 215 | 215 | 145 |

30,000 | 40,000 | 235 | 235 | 155 |

40,000 | 50,000 | 250 | 250 | 165 |

50,000 | 60,000 | 260 | 260 | 175 |

60,000 | 70,000 | 270 | 270 | 185 |

70,000 | 80,000 | 280 | 280 | 190 |

80,000 | 90,000 | 295 | 295 | 195 |

90,000 | 100,000 | 300 | 300 | 200 |

100,000 | 200,000 | 375 | 375 | 250 |

200,000 | 300,000 | 450 | 450 | 300 |

###### § 555.220 Table of distances of ammonium nitrate and blasting agents from explosives or blasting agents

Ammonium nitrate, by itself, is not considered to be a donor when applying this table. ammonium nitrate (AN), ammonium nitrate-fuel oil (ANFO) or combinations thereof are acceptors. If stores of AN are located within the sympathetic detonation distance of explosives or blasting agents, one-half the mass of the AN is to be included in the mass of the donor.

*Use the table at § 555.218 to determine required minimum distances from inhabited buildings, passenger railways, and public highways.*

### Fireworks

**Requirements for display fireworks, pyrotechnic compositions, and explosive materials used in assembling fireworks or articles pyrotechnic (excluding those in the process of manufacture, assembly, packaging, or transport).**

No more than 500 pounds (227 kg) of pyrotechnic compositions or explosive materials are permitted at one time in any fireworks mixing building, any building or area in which the pyrotechnic compositions or explosive materials are pressed or otherwise prepared for finishing or assembly, or any finishing or assembly building. All pyrotechnic compositions or explosive materials not in immediate use will be stored in covered, non-ferrous containers.

The maximum quantity of flash powder permitted in any fireworks process building is 10 pounds (4.5 kg).

All dry explosive powders and mixtures, partially assembled display fireworks, and finished display fireworks must be removed from fireworks process buildings at the conclusion of a day's operations and placed in approved magazines.

#### § 555.222 Table of distances between fireworks process buildings and between fireworks process and fireworks non-process buildings

Net weight (pounds) of fireworks, i.e. all pyrotechnic compositions, explosive materials and fuse only | Display fireworks (feet) - barricaded double distance if unbarricaded | Consumer fireworks (feet) - process buildings where consumer fireworks or articles pyrotechnic are processed |
---|---|---|

0-100 | 57 | 37 |

101-200 | 69 | 37 |

201-300 | 77 | 37 |

301-400 | 85 | 7 |

401-500 | 91 | 37 |

Above 500 | Not permitted | Not permitted |

#### Fireworks Process Building

*While consumer fireworks or articles pyrotechnic in a finished state are exempt, explosive materials used to manufacture or assemble such fireworks or articles are subject to regulation. Fireworks process buildings where consumer fireworks or articles pyrotechnic are being manufactured or processed must meet table of distance requirements.*

*A maximum of 500 pounds of in-process pyrotechnic compositions, either loose or in partially-assembled fireworks, is permitted in any fireworks process building.*

*Finished display fireworks may not be stored in a fireworks process building.*

*A maximum of 10 pounds of flash powder, either in loose form or in assembled units, is permitted in any fireworks process building. Quantities in excess of 10 pounds must be kept in an approved magazine.*

#### § 555.223 Table of distances between fireworks process buildings and other specified areas

Net weight (pounds) of fireworks, i.e. all pyrotechnic compositions, explosive materials and fuse only | Display fireworks (feet) | Consumer fireworks (feet) - process buildings where consumer fireworks or articles pyrotechnic are processed |
---|---|---|

0-100 | 200 | 25 |

101-200 | 200 | 50 |

201-300 | 200 | 50 |

301-400 | 200 | 50 |

401-500 | 200 | 50 |

Above 500 | Not permitted | Not permitted |

**When calculating the distance from passenger railways, public highways, fireworks plant buildings used to store consumer fireworks and articles pyrotechnic, magazines and fireworks shipping buildings, and inhabited buildings:**

*This table does not apply to the separation distances between fireworks process buildings (see § 555.222) and between magazines (see tables at §§ 555.218 and 555.224).*

*The distances in this table apply with or without artificial or natural barricades or screen barricades. However, the use of barricades is highly recommended.*

*No work of any kind, except to place/move items other than explosive materials from storage, may be conducted in any building designated as a warehouse. Fireworks plant warehouses are not subject to §§ 555.222 or 555.223.*

#### § 555.224 Table of distances for the storage of display fireworks (For bulk salutes, use table at § 555.218)

For the purposes of applying this table, the term "magazine" also includes fireworks shipping buildings for display fireworks.

Net weight (pounds) of firework, i.e. all pyrotechnic compositions, explosive materials and fuse only | Distance between magazine and inhabited building, passenger railway, or public highway (feet) | Distance between magazines (feet) |
---|---|---|

0-1000 | 150 | 100 |

1,001-5,000 | 230 | 150 |

5,001-10,000 | 300 | 200 |

Above 10,000 | Use Table § 555.218 |

For fireworks storage magazines in use prior to March 7, 1990, the distances in this table may be halved if properly barricaded between the magazine and potential receptor sites.