Basic Math Study Guide. MTI Learning Center Mitchell Training, Inc.

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1 Basic Math Study Guide MTI Learning Center 2001 Mitchell Training, Inc. 1

2 Basic Math - Section 1 Basic Math The math in this course is connected with solving equations. An equation is a set of numbers arranged on either side of an equals ( = ) sign. An equation is like a balance scale. To be true, the value of what is on one side of the equal sign has to be the same as what is on the other side. 2 = 2 and = 2 are true equations. This course is a review of the basic concepts involved in using equations in solving problems. We most often use the term formula to designate an equation commonly used to solve familiar problems. First, let s discuss the "transition" from spoken or descriptive language to mathematical language that can be used in calculations. A tank that is half-full is usually said to be just that, "half-full." That is not something we can put into a calculator, however, or enter into an equation. The fact is, we tend to talk in fractions which in this case would be written as "1 / 2." For the purposes of solving the problems, all amounts, distances, time, and measurements must be in decimal form. You must convert the language of measurement into decimals. The prefix "dec.." refers to ten. Our mathematical computations done on a common calculator are done in Base 10 meaning that the system uses 10 numbers to designate parts of a whole. We commonly use the term "decimal" to refer to a number which is a combination of whole parts and parts of ten. The first number to the right of the decimal point (".") is the tenths, the second number the hundredths, and so on. The number 45.67, we call a "decimal number," means that you have 45 whole units of something and 67 hundredths of one of those same units. So a tank that is half-full would be described in base 10 or decimal form as.5. How many feet is the diameter of a 6 inch pipe? The answer is.5 feet. The language of measurements is complicated by the various systems we have to use, such as pounds (base 16), feet (base 12). All metric measurements are already in base 10, making conversions between measurements very simple. 2

3 One-half a pound is 8/16, or.5. One-half a foot is 6/12, or.5. A nine-sixteenths wrench is 9/16 of an inch, or.56 of an inch. Converting common language into calculable numbers usually means converting from fractions to decimal numbers. Here s where many careless mistakes are made. Remember to enter the top number into the calculator first and then the operator and then the bottom number. Clear, enter 1, enter, enter 2, enter = This may seem like an oversimplification of a basic concept. However, it is the basis for all conversions, most problem solving, and transitions from language into mathematical calculations. You can convert any given measurement or amount into a decimal equivalent by this process. Percentages are merely a representation of a fractional amount of 100. "Per" means for each and "cent" is the term meaning 100. Twenty-five percent or 25% means 25 parts out of a hundred or 25 /

4 Basic Math Concepts Math is governed by a set of rules and principles which help to establish order and designate what you do and when you do it. For example, the most basic functions are: addition - shown by the symbol "+" subtraction - shown by the symbol "-" multiplication - shown by the symbol "x" or * division - shown by the symbol " " or / In an equation, these functions must be done in a certain order for the result to be correct. Multiplication and Division must be done first, with Addition and Subtraction being done last. There are ways you can remember this order, just make up a little jingle to help you remember, like: or M D A S "Mama Danced All Summer" You can calculate a string of multiplication and division functions or a string of addition and subtraction functions without any regard as to the order. For example, is exactly the same result as The function or sign appearing in front of any number designates what is to happen to it and goes with it. Such as Another rule of order involves the use of the parentheses ( ). When used in a formula, the computation inside the parentheses must be done first, according to the MDAS rules. Here is an example: 4 x (1 + 2) Even though the multiplication function is normally done before the addition, in this case the addition inside the parenthesis must be done first. 4 x (1 + 2) = 4 x 3 = 12 4

5 Now calculate these numbers: 4 (6-2) = = If you got "1" you did it right. You will sometimes see a number which has a superscript (called an exponent) which looks like this: 5 2. This little number above and to the right means that the main number (in this case 5) is to be multiplied by itself the number of times of the superscript number or exponent (in this case 2). This can be shown as 5 x 5. The number 5 3 can be shown as 5 x 5 x 5. 5

6 Fractions A common way of showing a part of a whole is by the use of fractions. As you know, most people use fractions when describing a part of something, such as "one-half of a load," "three and three-sixteenths of an inch," "four and a half gallons," and so on. Fractions are written in just that manner...one-half of something is written as 1 / 2, four and a half gallons is written as 4-1 / 2. Notice that whole numbers and fractions can be combined together. A fraction is simply a mathematical relationship between two numbers, one on top and one on the bottom, separated by a line. 1 or 1 / 3 3 The top number is called the numerator and bottom number is called the denominator. The bottom number tells us how many parts (or pieces) it takes to make a whole unit. In the above fraction, that number is 3. The top number tells us how many of these parts we actually have, in this case, 1. The fraction means that we have one out of three parts (or pieces) represented. You already know that fractions can be combined with whole numbers to form a number which looks like this /3. This number means that we have 14 whole units and 2 out of 3 parts of that same unit. 23-1/2 23 whole pieces and one-half of another 5-1/4 5 whole pieces and one-quarter of another 9-2/3 9 whole pieces and two-thirds of another To change a combination fraction into one fraction, just multiply the whole number by the number of parts in 1 whole number, that is the bottom number, add the top number, put that result on the top and you will get a fraction which is a combination of the whole number and the fraction: 5-1 / 4 = 5 times 4 which is 20, adding 1 making 21 / 4 (21 over 4). In an equation, having letters or numbers over and under a line (a fraction) means to divide the bottom number into the top number. A = B or A = B / C C means: A = B C 6

7 Decimals Like fractions, decimals can be used to represent pieces of whole numbers. In almost all cases, problems will be solved using decimals and the decimal equivalent of some fraction. Remember, fractions and decimals are so closely related that any fraction can be changed to a decimal simply by dividing the bottom number into the top number. Examples are: 1 / 2 =.5 12 / 16 =.75 4 / 12 =.25 Now, if any fraction can be converted into a decimal, how about the combination numbers like 14-1 / 2? First, we don't have to do anything to the number 14. It is already a whole number and doesn't need to be converted. What we must do is convert the 1 / 2 and add the result to the 14 to have a decimal which is the exact same thing as the original. Second, the conversion of the fraction is done like all other fractions, in this case just divide the bottom number (2) into the top number (1). With your calculator, do the division and you will come up with.5, which actually represents 50 parts out of a hundred, which is the same as 1 part out of two. Now, write the whole number and the decimal, separated by the decimal point and you get 14.5 You can convert the combination number into one fraction by multiplying the bottom number of the fraction part (in this case, 2) times the whole (in this case, 14) and adding the top number (in this case, 1) to make the numerator (top number) and placing the result over the bottom number (in this case, 2). 2 * turns out to be the fraction 29 / 2 or Rounding off is a term used to shorten long decimals to workable size. In most cases, you will only need one or two decimal places for accurate measurements on your job. Here's how you round off: becomes becomes 4.7 7

8 Here's the rule for rounding off to two places: If the next number (the third number) is 5 or higher, the second number is raised. If the third number is 4 or less, the second number remains the same. Rounding off: becomes becomes becomes becomes 1.5 (you don't need the last zero) In this course, you will find problems and answers rounded off to various levels. In come cases, it is traditional to work with a number rounded off to one or two or more places, such as the gallons per cubic foot of water, which is usually written as 7.48, even though 7.5 would work just as well. Remember, there is no correct way, only practical usage and specified levels of accuracy in problems. 8

9 Percentage Percentages (%) are another way of expressing parts of a whole. Actually, fractions, decimals and percentages are different ways of writing the same thing. Percentage is nothing more than expressing parts of a whole as so many parts out of 100 parts. If you visualize a whole unit being divided up into 100 equal parts, and if you take half of them, you have taken 50 parts, or 50% of the whole. Any number expressed in a decimal can be made a percent by moving the decimal point two (2) places to the right and adding the percent sign (%). Any percentage can be made into a decimal by moving the decimal point two places to the left and dropping the percent sign (%). Let s see how that works. Here are some numbers with decimal places. Note how the decimal point moves two places to the right to make the number a percent..65 becomes 65%.535 becomes 53.5%.5 becomes 50% (notice the added zero) 3.6 becomes 360% (add the zero to make 2 places) The added zeros are called place keepers and are very important in keeping the number accurate. Let s make some decimals out of some percentages: 54% becomes.54 12% becomes.12 2% becomes.02 (place keeper zero added) 125% becomes 1.25 What happens when a number like 4 is to be converted into a percent? Well, you must locate the decimal point just to the right of the number, then add place keeper zeros and move the decimal two places to the right. 4 (a whole number) 4. (locate the decimal point) 4.00 (add the place keepers) 400% (move the decimal point two places to the right) (If you have 4 wholes, the percent would be 400%) When converting a decimal to a percentage, always move the decimal place two places to the right, even if the numbers are already zeros: 9

10 .0005 becomes.05% (notice the decimal point is still there, even when the number is expressed as a percent. This number is not 5%) A common way of asking for a certain percentage is to state: "What is the percentage OF something IN something?" The use of the term "OF" means multiply. So a percentage of something needs a formula to show the multiplication of those somethings. Also, another way it is commonly asked is: "Something (a number) IS some percentage OF what number? Here is the formula: A percentage OF a number IS something: P x N = R or What IS some percent OF something? R = P x N In both cases the formula for percent is: R = PN N is the original number P is the percentage R is the result Want to figure out any percentage? Just substitute the values you know for the corresponding letters in the formula. Problem: What is 2 percent of 55 gallons? R = PN N = 55 gal. P = 2% R = 55 x.02 R = 1.1 gal (the formula) (the total amount) (the percentage you're looking for) (convert the percentage to a decimal) 10

11 Fraction/Decimal/Percent Exercises (Solutions found on Section 1 anwer sheet) Fraction Convert these fractions to decimals: Decimal 1) 3/4 2) 4 3/8 3) 10/12 4) 16/12 Convert these percentages to decimals: Percent 5) 344% 6) 12% 7) 34.5% 8) 78.50% 9).67% 10).04% 11).0087% Convert these decimals into percents: 12) ) ) ) ).46 17) ) End of Learning Event 1 11

12 Basic Math - Section 2 Using Basic Algebra Here is a review of the basic concept of algebra and how it is used to solve equations. An equation just means that something on one side of an equal (=) sign is the same as what's on the other side. Formulas are used in general terms, showing the units of measure which are to be used and what is to be done with them. In some problems, formulas appear which have letters in them as well as numbers. Algebra in its basic form is about using symbols to represent values. The traditional way of expressing an unknown value is the letter X. If symbols are to be used in equations, the letter X always means an unknown value and not the multiplication sign. A formula generally contains symbols to represent values. Any equation must be true, therefore any symbol on one side of the equal sign must represent the same value as the symbol on the other side. A = B means that whatever A is, B is also. A = B + C means that A is the same value as B and C added together. When two symbols are shown together, it generally means to multiply them together. A = BC would then mean that B and C are multiplied to get the value of A. Sometimes in writing equations, the star (*) is used to show multiplication (A = B * C). The same is true of a number and a symbol shown together, such as 4.5 L, which in an equation means to multiply 4.5 times L or 4.5 * L. You can see how the X or x used to show multiplication might be confusing. Substitution is the term used to show what is done about those letters which are popping up in the formulas. It s a simple matter in solving problems with formulas to substitute numbers for the letters and then do the calculations. In the problems used in this class, there must always be a value given for all but one of the symbols in order to do the calculations and solve the problem. Now we can substitute numbers for the letters and work the problem. In this case, the formula A = LW is the formula for finding the area of a rectangle: Area is equal to the Length of the rectangle times the Width is the way you would say this problem. 12

13 For now, we will be using the same units of measurement in the problem, such as feet, yards, pounds, etc. Later, you will learn how to covert units so that the formulas can be worked using the same units. You can't add feet to gallons or multiply meters by inches, and so forth. So, let's work this formula. We are trying to find the area, let's say, in square feet. To find that area, you multiply the length of the rectangle times the width. A (area) = L (length) x W (width) The length is 12 feet and the width is 4 feet. Now, let's substitute the numbers for the letters: A = 12 x 4 A = 48 If the units of measure were feet, as they are in this case, the answer will be in square feet, so: the area of the rectangle equals 48 square feet, or 48 sq.ft., or 48ft 2. This is how all formulas are worked. Substitute the numbers for the letters and do the calculations. In the space below, solve the formula: A = BC where B = 6 and C = 2.5 A =? If you got 15, you got the correct answer. You don't know what units the 15 represents because there were no units give in the problem. There are many ways to write formula solutions as you work them. You can keep everything straight if you write down every step as you work the problem. A = LW A = 4 x 5 A = 20 The known substitutions are not always on the same side of the equal sign. When that happens, you must remember that both sides of the equation must be the same value, so if you add, subtract, multiply, or divide to one side, you must do the same thing to the other side. Take the equation A = LW. Suppose we had the area (A) given and one side or the length (L) and wanted to know the value of the other side. The substitutions would look like this: 13

14 A = LW A = 23, L = 4 then 23 = 4W (When a number and a symbol are placed together it means to multiply) 23 = 4 * W What you must do is then get the equation to have only the unknown, in this case, W, on one side and the numbers on the other. To do that, you can do something to both sides, something which will leave only W on one side. If you divided 4W by 4, you would have 1W, or just W which is what you want. So, divide both sides by 4, writing it something like this: 23 / 4 = 4W / 4 or 23 / 4 = W 5.75 = W which is of course the same as W = 5.75 Now solve the equation if the numbers are: A = 235 L = 23.5 W =? Answer is W = 10 Sometimes an equation will have what is apparently a fraction on one or both sides. Solving becomes a matter of reducing all the information on each part of the fraction(s) before solving the equation for 1 unknown. Here is a formula with 4 symbols or unknowns, three of which must be substituted for solving. G = MP ( M = 2.3, P =.56, and D = 4) D G = 2.3 *.56 G = G =.32 Suppose the unknown is on the other side and is part of the fraction...32 = 2.3P 4 There are several tricks to solving, all of which involve remembering whether to multiply or divide and what into what and what by what. A simple way to get it right is to use the principle of the ratio. A ratio is usually expressed in terms of two fractions which are equal in value, such as 1 / 2 and 5 /

15 1 = Our common knowledge about fractions tells us that these two are equal. However, if there happens to be an unknown anywhere in the equation, it must be solved by cross multiplication. 1 = D 2 10 The result is an equation 2 * D = 1 * 10 2D = 10 D = 10 / 2 D = 5 Back to the equation..32 = 2.3P 4 You can make a ratio-like equation by making a fraction on the non-fraction side of the equal by placing a 1 under the number, (dividing by one does not change the value).32 = 2.3P 1 4 Cross multiply 1 * 2.3P =.32 * 4 2.3P = 1.28 P = 1.28 / 2.3 P =.56 You can prove the solution by substituting all the numbers and doing the calculations..32 = 2.3 * = =.32 15

16 Here is where you can punch the numbers into the calculator in a string: 2.3 x.56 4 = or 2.3 *.56 / 4 = Solve this equation: V = D (V = 2.3, T = 180) T D =? Answer is D =

17 Conversions At this point, you need to know how to convert units of measurement to other units of measurement. As you saw earlier, you cannot multiply inches by yards and get an accurate answer. There are several common units with which you must work, such as distances, time, temperature, and so on. One example would be in finding the volume of a trench measured in feet and having to know the amount of cubic yards of material to be hauled off. To change from one unit to another requires the use of a conversion factor which relates the two together. Most of the time, these factors are found in charts, but they are not always practical for on the job use and are difficult to remember accurately. Here is a way of converting units using ratios. Here is the way a ratio is stated: A is to B as C is to D Not as complicated as it sounds. Actually, a ratio is a way of using two equal fractions to find an unknown. We know that any number is the same as that number divided by 1. So, any fraction which has a 1 on the bottom has the same value as the number on the top. The line between the numbers means to divide, so any number divided by 1 is the same as the number itself! Look at these two fractions: Now, you can tell that they have the same value, proven by converting each into a decimal. That decimal being.5, or just another way of saying a half. We can say that the two fractions are equal, or that 1 is to 2 as 2 is to 4. The language of the ratio. 1 = Suppose that you only know three of the numbers. Well, you use a symbol, like X, to take the place of the unknown. 5 = _X ,000 This shows that there are 5 parts out of 234 and you want to know how many parts out or 10,000 would make the equation true. X represents a number that 17

18 would be the same ratio of parts as 5 in 234. You could also say that the percentages would be the same. You solve this equation by cross multiplying: then: 5 = X , X = 5 * 10, X = 50,000 X = 50,000 / 234 X = Remember, you must make any conversions before you substitute the units (numbers) into a formula. Here's another example. Change 4.5 hours into minutes. (You already know that 4.5 hours is 60 minutes times 4.5. However, working this problem using the conversion technique will help you to be able to solve more complicated problems in your job.) First, write the fraction for minutes in an hour ( 60 min is to 1 hr ) 60 min. 1 hr Now write the fraction with the unknown ( as X is to 4.5 hours ) then: X 4.5 X = X = 60 * 4.5 = 270 min Practical Conversion Hints: When to multiply? When to divide? What by what? It can get confusing, even to a person who works with figures all the time. For example, if converting feet to yards in a problem, do you divide by 12 or multiply by 9 or divide by 3 or what? Here s a good rule to remember: 18

19 First, if you are unsure, write the numbers down with the single unit listed first, like 1 = 12 or 1 gal = 8.34 lbs. Then start with the measurement you are given, like 467 gal: 1 gal = 8.34 lbs. and underneath write: 467 gal =? lbs. When going from left to right, always multiply. So you would write: 467 * 8.34 Now suppose you have 4359 lbs. and you want to know how many gallons that is. Write it down: 1gal. = 8.34 lbs? = 4359 lbs When going from right to left, always divide. So you have: 4359 / 8.34 Common conversion factors: 1 ft = 12 inches 1 yd = 3 ft 1 cu.ft. = 1728 cu.in. 1 cu.yd. = 27 cu.ft. 1 cu.ft. = 7.48 gal 1 sq.yd. = 9 sq.ft. 1 cu.ft. = 62.4 lbs 1 gal = 8.34 lbs Converting common measurements into decimals, as shown in previous sections, involves creating a fraction which describes the measurement in its own language and units. Examples are inches / feet feet / yards eighths / inches sixteenths / inches ounces / pounds These fractions will produce a decimal conversion which can be calculated in a base 10 calculator. 19

20 Metric Measurements The metric system uses a meter as the basic unit of length measure, a liter as the basic units of volume and weight. A meter is about the same length as a yard, and if you want to know how many inches make up a meter, you can look it up in a chart or use a ratio. In most cases, there is no practical need to convert into metric measures as long as you are converting to base ten or the decimal system. The metric system is already in base 10 and can be calculated directly with a calculator. Here are some common metric measurements and the decimal equivalents: 1 meter 1 1 decameter.1 1 centimeter.01 1 millimeter liter (weight) 1 1 gram milligram liter (liquid measure 1 1 deciliter.1 1 centiliter.01 20

21 How To Do Word Problems Word problems are the basis for most technical exams and the solution for most all practical problems. A word problem is a technical problem stated in everyday terms. Most problems start that way, anyhow. This section will describe word problems, tell you how to solve them, and present some practice problems for learning activities. Let's say you want to know how much dirt will be removed from a pit. You will ask these questions: "How big is the hole?" "How much material will be removed?" What will it take to haul the material away?" The word problem would be stated this way: "How many truck loads of material will be removed from a pit 20 feet long and 10 feet wide by 6 inches deep if the trucks will haul 6 cubic yards of material at a time?" Step 1: Identify the problem. This is obviously a problem of the volume of a cube. Additionally, the information will have to be converted into cubic yards to determine the truck loads. Step 2: Determine the mathematics needed and the formula(s). In this case, the volume formula of length times width times height will be used. Step 3: Decide on the measurements to be used. We will need to know the cubic yards so a determination will have to be made to see if all the calculations can be made in yards, or if conversions will be needed. You know from the table or from memory that there are 27 cubic feet in a cubic yard. If you make the measurements in the unit which is called for in the problem, you won t have to remember any conversion after the calculation. In this case, you would convert everything into yards before substituting the numbers into the formula. Step 4: Determine what degree of accuracy will be needed. Filling up a truck with dirt is not very exact. Therefore, the degree of accuracy will not be great, perhaps rounding off to the nearest whole number. 21

22 Step 5. Write down the figures and the formula(s) and do the calculations. In this case you would write down: Substitute: V = LWH V = 50 ft x 40 ft x.5 ft (6 in. is 6/12 or.5 ft.) = (the dimensions of the pit in ft) V = 1000 ft 3 (the answer in cubic feet) Convert: There are 27 cubic feet in a cubic yard, therefore to find out how many cubic yards of material there are, just divide 27 into the total amount of cubic feet (1000) = 37 yd 3 (actually rounded down) 27 Each truck hauls 6 yd 3 37 = 6.1 (looks like there will be a little over 6 truck 6 loads) An alternate conversion would have been to convert to yards before the calculation. You sometimes have different measurements to consider, such as a combination of feet and yards or feet and inches. For example: 50 / 3 = 16.7 yd. (gives you the conversion of 50 feet to yards ) 40 / 3 = 13.3 yd. (gives you the conversion of 40 feet to yards) 6/36 =.17 yd. (gives you the conversion of 6 in.. to yards) * 13.3 *.17 = 37.7 (accounting for rounding, the same as before) Review of steps in solving problems: Step 1 - Identify the problem Step 2 - Find the formula Step 3 - Choose the measurements (and conversions) Step 4 - Determine the accuracy needed Step 5 - Do the calculations 22

23 Equations/Conversions/Decimal Exercises (Solutions found on Section 2 answer sheet) Make the substitutions and solve these equations: For all equations use these values: Equations A = 3.5 B = 34 C = 2 D =.46 1) X = AB 2) X = ABC 3) X = 5 / D 4) A = D / X 5) AX = CD 6) 456A = X Conversion Make these conversions: 1) change 45.3 ft 3 into gallons 2) convert 2 ft 3 into pounds 3) change 45 ft 2 into yd 2 4) change 3.5 gallons per second into gallons per hour 5) substitute the numbers in the equation A = LW if L = 45 yd. and W = 4 ft Decimal Make decimals from these measurements: 1) 23 inches = feet 2) 6 feet = yards 3) 9 sixteenths = inches 4) ounces = pounds End of Learning Event 2 23

24 Basic Math - Section 3 Area Areas are flat surfaces such as desktops, floors, and walls. They have two dimensions only, lengths and widths, or lengths and heights depending upon what type of surface is being measured. The formula used to calculate the area of a square or a rectangle is written: A = LW (Area = Length times Width) What is the area of a concrete floor that measures 10 feet wide and is 25 feet in length? First of all, pick out the key words. It's not important whether it is a concrete floor or a driveway. What is important is that it is an area. The shape is a rectangle, a length and a width are given, and all the dimensions, or units, are in feet. With this information, you can choose the correct formula to use, set up the problem, and solve it easily. You should always draw a simple diagram where possible, which will more clearly illustrate what you are trying to solve. If the dimensions are given in different types of units, such as length in feet and width in yards, the conversion to common units can be made right on the diagram. L = 25 W = 10 Never multiply different units, such as feet times inches, or yards times miles. Do all the conversions to common units first, and then bring the common units into the formula. This will be illustrated in later examples. 24

25 Solving the problem: A = LW A = 25 ft. x 10 ft. A = 250 sq. ft. Notice that the answer is in "square feet, written as sq.ft. or ft 2. Remember that the answer to any area problem will be in "square" units. Again, if the dimensions are given in different units, you must convert to "common units" before entering any numbers into a formula. Notice, too, that we call some rectangles squares, just to denote that all sides are equal. The other shape that you will encounter often is the circle. The parts of a circle are: Circumference: The distance around the circle. Diameter: A straight line drawn through the center of the circle from one side to the other. Radius: A line drawn from the center of the circle to one edge.. The radius equals 1/2 the diameter. Circumference Radius Diameter The traditional formula for the area of a circle contains a term called pi. It is written mathematically as a symbol that looks like this: π and it is equal to The formula for the area of a circle is: A = π R 2 (Area equals pi times the Radius times the Radius) What is the area of a circle that has a diameter of 50 feet? 25

26 First, draw a simple diagram. You can see that the diameter is given, but you need the radius. R = 50 ft. / 2 = 25 ft. A = πr 2 A = 3.14 x (25 ft. x 25 ft.) A = 3.14 x 625 A = ft 2 (The 25 ft. is the radius, or one-half of the diameter. Remember that any number "squared" is that number multiplied by itself. So, feet times feet equals square feet.) Many errors are made in hasty calculations by forgetting to use the radius instead of the diameter. Sometimes the radius is the given dimension but more often the diameter is given, such as the size of pipe or the size of a tank. (You ve never heard of 5 radius pipe, it s always given as 10 pipe by definition.) Actually, the best way to find the area of a circle when the diameter is the given dimension is with this method. Use the formula.785d 2. Here s the source of the method. The diameter of a circle, or the distance across at the widest point is the same as the length or width of the rectangle (square) of the same size (see the diagram below). Now, if we multiply the length times the width, we are using the same measurement as the diameter of the circle but we get the area of the square, not the circle. However, the area taken up by the circle in a square of the same dimensions is always the same ratio regardless of the size of the square and corresponding circle. So we have to make an adjustment to leave out the areas in the corner which are not part of the circle. Better still, we ll go right to the area of the square that the circle occupies. L W D The area of the circle is always 78.5% (.785) of the square. If the sides of this square are 12 inches, then the diameter of the circle is twelve inches. If we multiply 12 x 12, we get an answer of 144 square inches, which is 26

27 the area of the square. If we multiply the Diameter times the Diameter (D 2 ), we get 144 square inches, which is definitely not the area of the 12 inch circle. The circle actually takes up 78.5% of the square. If you take 78.5% of 144, you will get the area of the circle! So, now we have a simple way to find the area of the circle, just multiply the diameter by the diameter and then multiply by 78.5% or.785. The formula looks like this: A =.785D 2 or.785(d 2 ) or.785 * D * D Either of those will work. You can enter the numbers into the calculator easily in a string, the.785 does not have to be entered first, but it helps to get the equation started right. Here is an example: Let s find the area of a circle with a diameter of.5 feet (6 ). (Sounds like the end of a 6 pipe.) A =.785(D 2 ) A =.785 *.5 *.5 A =.196 A =.2 (rounded off to one place) So the area of the end of a piece of 6 pipe, which is a circle, is.2 sq. ft. ( ft 2.)! End Area 6 =.2ft 2 So far you have learned to calculate the areas of the two basic flat surface shapes. There is one more formula that you need to become familiar with that deals with circles. If the distance around a circle needs to be calculated (the circumference), the formula is as follows: C = π D (Circumference equals pi times the Diameter) 27

28 What is the length of a strap around a 50 ft diameter tank? You can see that we are looking for the circumference of the tank, so you can choose the correct formula. The diameter is given, so C = π D C = 3.14 X 50 ft. C = 157 ft. Tank circumference = 157 ft. 28

29 Volume of Box Shapes and Cubes Remember that areas are two-dimensional. Volumes are three-dimensional. The rectangular shapes have lengths, widths, and heights (or depths), where the circular or "cylindrical" shapes have heights (or depths and sometimes lengths) and diameters. * Sometimes you may wish to use C for Capacity instead of V for Volume. This might avoid confusion with another calculation, used to find the Velocity, where the traditional symbol is V. Some of the formulas in this manual will use Capacity so you can become accustomed to its usage. Volumes are measured in "cubic" units. If the units were feet, it would be cubic feet, or cu. ft. or ft 3. The formula for calculating these shapes is written: V = LWH (Volume = length * width * height) Notice that there are three dimensions: length, width, and height. Sometimes these dimensions will be given in different units in one problem, such as a length in yards, height in inches, and width in feet. Again, remember to convert to common units before entering any numbers into the formula. What is the volume of a rectangular tank that measures 20 feet long, 15 feet wide, and 10 feet high? Now use the steps. Remember to pick out the key words. They ask for volume and give dimensions. 15 ft 10 ft 20 ft 29

30 In volume problems, it might be less confusing if you can draw a diagram of the object in question. Label the dimensions, and if any conversions are required, do them directly on the diagram. V = LWH V = 20 ft. x 15 ft. x 10 ft. V = 3, 000 cu. ft. Volume = 3, 000 cu. ft. 30

31 Volume of Cylindrical Shapes Just as with areas, you can use either volume formula for all cylindrical shapes. The units for cylinder volumes will be "cubic" units, just as in the rectangular shapes. This means that there are three measurements for each shape, just like a cube or box. In a cylinder, the height or length measurement is the barrel or tube part of the shape. L D In calculating volumes for these shapes you can use either formula: V =.785D 2 L or V = πr 2 H or (V = AL, which means area times length) In problems involving volumes of pipes, the most common mistake is not using the same "units" in the substitution step of the problem. Usually, pipe diameters are given in inches and sometimes pipe lengths are given in miles. Make sure that you convert these into common units before the substitution step. Usually, feet will be the units used. It is good practice to use ft. measurements for all pipe areas. To convert inches to feet - I/12 To convert miles to feet - M * 5280 To convert yards to feet - Y * 3 To convert feet to yards - F / 3 Calculate the volume of a cylindrical storage tank that has a diameter of 24 feet and a height of 25 feet. V =.785 D 2 L V =.785 x 24 ft. x 24 ft. x 25 ft. V = 11,304 cu. ft. Using the traditional formula π R 2 L will give you the same answer. 31

32 A pipeline has a diameter of 8 inches and runs a distance of 4.5 miles. What is the volume, in cubic feet, of this pipeline? A good practice is to draw a diagram, entering the converted dimensions. Learn to do these types of problems in feet, yielding a useful result. L = 4.5 mi. (4.5 x 5,280) = 23,760 feet D = 8 inches (8 / 12) =.67 ft. Calculate: V =.785 D 2 L V =.785 x.67 ft. x.67 ft. x 23,760 ft. V = 8,372.7 cu. ft. 32

33 Area / Cube Volume / Cylinder Volume Exercises (Solutions on Section 3 answer page) Areas 1) What is the area in square feet of a rectangle that measures 16 inches in length by 2.3 feet in width? (hint: remember to convert inches to feet--16/12) 2) What is the surface area of a tank that is 46 feet in diameter? 3) What is the area of the shaded area of this diagram? (Could this be a driveway?) ) What is the area of the inside wall of a tank 46 feet in diameter and 8 feet high? Cube Volume 5) How many cubic feet of water will be needed to fill a tank that measures17 ft. by 11.5 ft. by 42 inches? How many gallons? 6) How many cubic yards of dirt will be removed from a ditch that is 5 feet wide, 4.5 feet deep and 1 mile long? Cylinder Volume 7) What is the volume of a 16 inch pipe that is 10 feet long? What is its capacity in gallons? 8) What is the volume of a tank that has a diameter of 46 feet and is17 feet deep? 33

34 9) What is the volume, in cubic inches, of an 8 inch cylinder that is 5 feet long? 10) What is the volume of a 10 inch pipe that is16 feet long? Conversions 11) How many gallons in a 10 ft. deep tank with a diameter of 46 ft.? (hint - remember the conversions in the last section, start with the measurement you are given) 12) Convert these pipe sizes to feet: 4 = 6 = 8 = 10 = 12 = 16 = 18 = 24 = 30 = 36 = 40 = 48 = 54 = 60 = This chart will be useful in calculating the volume of any pipe size in feet. 34

35 Basic Math - Section 4 Velocity, Flow Rates and Retention Time Velocity, flow rates and retention time have one thing in common. These calculations all have an added dimension (other than shape or size) and that is time. In most cases, the calculations derive from a volume or mass moving at a specified speed for a given period of time. In driving, we use the term miles per hour, and most other technical applications use feet per second, two measurements easy to get. Velocity is the speed at which something is traveling, usually shown as a fraction such m / h (mph) or f / s (fps), the slash mark denoting per. Flow is a measure of speed as well as a specific amount of, say water, flowing at that speed. It is usually shown as gal. / min. (gpm), ft 3 / sec. (fps), gal. / hour (gph), or MGD (million gallons per day). It is used to determine the rate at which water is flowing through a pipe or other conduit, and the rate at which water is being pumped. Retention time (sometimes called detention time) is the theoretical time a given amount of water stays in a pond, tank, or other container. The calculated time is based on the flow rate into and out of the container (both being the same). Another way of stating this situation is to calculate how long it takes to fill or empty the container at a certain flow rate. 35

36 Velocity The velocity, or speed, of any moving object or mass is measured by the formula: V = D / T Velocity is the speed calculated in two units combining the distance traveled in a specified amount of time. The value is a combination of these two, and is usually stated in the form of distance per time. Some common units are miles per hour and feet per second and gallons per minute. If the calculation involves a unit like gallons, then the same principles apply, that is, a specified number of gallons per a unit of time. Do not be confused by the difference in volume and the capacity of containers or pipe calculated separately and converted into gallons, if necessary. D is the distance traveled T is the time it takes to travel the distance What would be the velocity of a train traveling 100 miles in 89 minutes? V = D / T (D = 100, T = 89 V = 100 / 89 V = 1.12 miles per minute. The answer may not turn out to be a workable number, given the uses for the information. You could, at this point, do a conversion to the useful unit, such as miles per hour. You could multiply the 1.12 times 60 to get miles per hour (1.12 * 60 = 67) miles per hour. A less confusing way is to do the conversions before using the formula. Suppose that the measured velocity was 3.5 f/s and the distance was 1 mile. How long would it take to travel that distance? V = D / T 3.5 = 5280 / T (feet in one mile) 3.5 T = 5280 (cross multiply) (3.5 / 1 = 5280 / T) T = 5280 / 3.5 T = 1,508 sec or 1,508 / 60 = 25.1 min To calculate the distance something might travel, you must have the time and the rate of travel. How much pipe can you lay in a week if you can lay 420 feet in one work day? 36

37 V = D / T 420 f/d = D / 7 days D = 2940 ft (cross multiply) 37

38 Flow Rate Determining how much water is flowing through a channel or pipe concerns the amount or quantity of water. Using a combination of the velocity and area formulas, the formula for flow rates is: Q = AV Q = quantity (or amount) A = area (area filled by the flow can be a full or partially filled pipe, channel or sluice of any shape. V = velocity (speed or time) A rectangular channel 3 feet wide contains water 2 feet deep and flows at a velocity of 1.5 feet per second. What is the flow rate in cubic feet per second? In this problem, an area is required. The width and depth of the channel are given, and when this is calculated, the resulting units are in square feet. Next, this answer is multiplied by the velocity, which is in cubic feet per second. Feet per second times square feet results in cubic feet per second. It may appear that two units are being calculated together. In a way that s true, however, the answer must reflect both the measurements. Q = AV (remember that the V stands for velocity) Q = (3 ft x 2 ft) x 1.5 ft./ sec. Q = 6 sq. ft. x 1.5 ft./ sec. Q = 9 cu. ft./ sec. Water in a 6 piece of pipe has a velocity of 2.5 ft./sec. What is the quantity of water passing through the pipe? Here s where the area calculation of ft. equivalents of pipe sizes comes in handy. You can calculate the ft 2 of any size pipe in the problem. In practical application, you remember or look up the areas of any size pipe. Applying the formula, you have: Q = AV (A =.785 *.5 *.5, or.2 ft 2 ) (V = 2.5 f/s) Q =.2 * 2.5 Q =.5 ft 3 ps (ft 3 per second) Remembering that you have three measurements involved (time and 2 dimensions will give you cubic measurements, in this case, ft 3.) 38

39 Partially Filled Pipe If the pipe or conduit is round and the water is not filling it, you can calculate the flow by calculating the area of the water as shown. Diameter of Pipe Water Line Depth of Water The calculation uses a ratio of the depth of the water in the pipe ( d ) to the diameter of the pipe (D). It does not matter what the units of measure are as long as they are the same (inches = inches, feet = feet, etc.). Inches are the most practical. Because the width gets increasingly smaller as the water gets deeper than half or greater than half, the calculation must use a function as shown below. Let s say that you have a 36 inch pipe and the water depth is 15 inches. (The only way you might know this is to measure it!) Let s also say that the velocity is 3 feet per second. (You would also have to measure this as well.) First, calculate the ratio: 1) d / D (depth divided by diameter) gives you a decimal, use 2 places 2) 15 / 36 =.42 (using inches) 3) Use the chart on the next page, find the d/d column and search down until you find the number you created in step ) Find the corresponding 4 decimal place number to the right of your number in the Factor column. 5) (don t mess up the decimal points). 39

40 6) Write the area for a circle formula and substitute the factor for.785 A =.3527 * D * D A =.3527 * 3 * 3 A = 3.17 ft 2 What you just calculated is the area of the water going through or coming out of the pipe, not the area of the opening of the pipe. Area of the water Now you can calculate the flow rate with the formula: Q = AV (A = 3.17 ft 2 and V = 3 ft per sec) Q = 3.17 * 3 Q = 9.51 ft 3 / sec With the calculation of flow rate, we can then determine the amount of water flowing in any measure we wish, such as gallons or millions of gallons. The time, measured in seconds, can be calculated to conform to any practical measure of time, such as minutes, hours, or days. To make accurate conversions, use the method of listing the given unit first and substituting the amount. 60 seconds = 1 min 60 min = 1 hour 24 hours = 1 day 1 ft 3 = 7.48 gal 40

41 Retention Time Calculating retention time is another flow rate problem with the addition that the volume of water being measured is not flowing through a uniform pipe or conduit but is a measure of an amount of water flowing into a container (or flowing out of a container). Actually, the calculation is intended to be a measure of how long an amount of water spends in such a container. Therefore, the only practical measurement that can be made is to calculate how much time it would take to fill (or empty) the container. An example of this type of application would be to know the retention time of chlorinated water in a storage tank to determine the chlorine residual necessary to have in the water so as not to lose too much of it over time as the water remains in the tank. Actually, this measurement is useful to determine the potential of pumping applications for moving water into or out of any type container, time being an important factor as well as amount of water moved. The formula used in these problems is the same no matter if it is a retention time, filling a container, or emptying a container. The formula for retention times is: R = Capacity of tank in gallons (volume in ft 3 times 7.48) r/t (Rate of flow in gallons per time unit) Retention time equals the capacity of the tank in gallons divided by the rate of flow in gallons per time unit. Notice that the volume of the tank is given as the capacity. In most cases, retention time problems will require you to calculate the capacity of the tank. This can be part of the formula or can be done in a separate calculation. Now, calculate this problem: How long will it take to fill a storage tank using a service pump rated at 600 gallons per minute if the tank capacity is 875,000 gallons? The tank capacity is given, so there is no need to calculate it. R = C (for gallons capacity) r / t (for rate / time unit) 41

42 R = 875,000 gallons 600 gal. / min. R = 1,458 minutes / 60 min R = 25 hrs The answer will reflect the units included in the measurements, in this case gallons and minutes. Conversion will be necessary if the answer must be in a unit different from the unit given (in this case, hours instead of minutes). If multiple pumps are used to empty or fill a tank, all the pump ratings will be added together. What is the retention time in a tank that measures 13 feet deep, 20 feet wide and 45 feet long if the flow through this tank is 450 gallons per minute? The shape is a rectangular box, so you know the formula to use from the volumes section. V = LWH V = 45 ft. X 20 ft. X 12 ft. V = 10, 800 cu. ft. C = 10, 800 cu. ft. x 7.48 gallons Capacity = gallons Now, you have calculated a key element in solving this problem, which is the tank capacity. The flow is given, so now write the formula and proceed. R = C r/t R = 80,784 gallons 350 gal. / min. R = minutes / 60 R = 3.8 hrs. 42

43 Velocity / Flow Rate / Retention Time Exercises (Solutions on Section 4 answer page) Velocity 1) It takes 2 minutes and 15 seconds for a raft to travel 285 feet. How fast is the raft moving? 2) A raft travels 325 feet at the rate of 3.5 ft./sec. How long did it take? 3) A raft travels for 3 minutes and 22 seconds at the rate of 4.5 ft./sec. How far did the raft travel? 4) How fast is a car traveling if it goes 650 miles in 5 hours and 30 minutes? 5) If you travel at the rate of 85 miles per hour for 4 hours, how far will you travel? 6) If it takes a float to travel 235 feet in 2 minutes and 15 seconds, what is the velocity of the water? 7) If a float travels 435 feet in 3 minutes and 35 seconds, how fast is the water moving? Flow Rate 8) Water in a 6 inch pipe is flowing at the rate of 3.5 ft./sec. Calculate the quantity in cubic feet. 9) How many cubic feet of water will flow from a 10 inch pump at the rate of 4.5 ft./sec? 10) A 16 inch pipe has a flow rate of 2.5 ft./sec. Calculate the number of gallons per second. 11) Calculate the number of gallons per minute flowing from an 8 inch pipe at the rate of 3.5 ft./sec. 12) At the rate of 2.75 ft./sec., how many gallons will flow from a 6 inch pipe? 43

44 13) Convert 2.5 ft 3 /sec to: gal per sec = gal per min = gal per hour = gal per day = 14) Convert 450,000 gal per day to: gal per hour = gal per sec = ft 3 per sec = Retention Time 15) What is the retention time for a 325,000 gallon tank that is being filled at the rate of 450 gallons per minute? 16) Calculate the retention time for a 495,000 gallon storage tank being emptied at the rate of 500 gallons per hour. 17) Calculate the retention time for a tank holding 55 gallons that is emptied at the rate of 3.5 gallons per minute. 18) A tank is 8 feet in diameter and 16 feet deep. The rate of flow into it is 3.6 gallons per second. Calculate the retention time in hours. 19) How long will it take to empty a pond using a service pump rated at 22 gallons per minute if the pond capacity is 350,000 gallons? 44

45 Basic Math - Section 5 Hydraulics Hydraulics is the study of liquids under pressure and/or in motion. The effect of water motion and pressure are essentially the same, force is exerted against some surface. Many uses for the energy created by hydraulic action are commonplace, including the generation of electricity and the hydraulic cylinder. Several principles of natural physics are at work; inertia, gravity, force, thrust, and pressure to name some. Mathematical calculations of these actions are discussed in this section. Inertia is the tendency of matter to stay in its existing state or motion. A mass of water behaves much like a solid object when confined in a pipe. Whenever there is a change of direction or motion, a condition will be created in the pipe that is connected with the energy potential of the water in the pipe. These forces can be damaging to structures is not properly prepared for. This energy is greater when the changes happen in a short time (rapidly), rather that over a long duration. For example, if you increase the pressure in a pipe gradually over time, the water will respond without dramatic results. However, if a sudden increase in pressure is applied, the forces are greater and the results will be sudden and drastic. If there is a spinning object, inertia causes force or pressure to be directed outward from the center of the spinning object in the direction away from the center. This is called centrifugal action or force. Force is the energy applied to a surface by an object either due to gravity or to pressure. Pressure is the same as force, in that it is what causes force. For example, if there is a container in which there is liquid under pressure, force is the energy exerted on the surface of the container. Gravity is the pull of a celestial body, such as the earth on objects. It is what keeps us falling off into space and what creates weight. Stuff with more mass is pulled with more force, thus weighing more. Gravity pulls on objects at the same speed, regardless of the mass. Even though things might weigh more (more mass) that other things, the speed at which they are pulled toward the center of the earth will be the same. If you dropped a ten ton weight from any height, and dropped a one ounce weight from the same height, they would strike the ground at the same time, providing the friction from the air was the same. This makes some calculations regarding hydraulics useful for practical purposes. 45