# Any Excuse for Math! Example T&I Applications Excel Tutorial – Read This FirstExcel Tutorial – Read This First Table of Contents Automotives – Ratio Automotives.

## Presentation on theme: "Any Excuse for Math! Example T&I Applications Excel Tutorial – Read This FirstExcel Tutorial – Read This First Table of Contents Automotives – Ratio Automotives."— Presentation transcript:

Ratios (Automotives, brakes, steering, etc.) Mechanical Advantage (MA) MA = Output Force / Input Force MA = Effort Distance / Load Distance MA = 400 lbs/100 lbs = 4 MA = 4”/ 1” = 4 You cannot get more out of a machine than what is put into it. With 100 pounds of force applied to a 1 sq.in. area cylinder, 400 pounds of force can be achieved at the output cylinder. This is because the output cylinder is 4 sq.in. in area. The tradeoff is that the output cylinder cannot move but one-fourth the distance of the input cylinder. Table of Contents Get the math teacher to supply you with similar problems.

Algebraic Expressions (Collision Repair, surface area) Surface Area Rectangle = lw (l = length, w = width) Cube = 6s 2 (s = side length) Rectangular Prism = 2(lw + lh + wh) Present students with the equation for a cube as: 6 (2w + 2l) Ask students to describe a cube. For a cube, width and length are the same. Let students know the most simple expression for surface area of a cube: 6s 2 Ask students to simplify the following expression. It will show that x and y are the same like w and l are the same for a cube. 6s 2 = 6 (2w + 2l) Substitute x for s and w they are the same, and let l be y to use standard variables. 6x 2 = 6 (2x + 2y) 6 continued Table of Contents

Algebraic Expressions (Collision Repair, surface area of a car, cont.) 6x 2 = 6 (2x + 2y) 6 X 2 = 2x + 2y X 2 -2x = 2x + 2y – 2x X 2 -2x = 2y 2 2 2 X 2 -x = y 2 Now see if x and y are the same. Let x = 4, and solve for y. y = 4 2 – 4 = 16 – 4 = 8 – 4 = 4 y = 4 2 2 Get the mathematics teacher to supply you with similar problems for students to practice in the abstract. Table of Contents In auto body class, it’s virtually impossible to determine the surface area of an automobile. However, in the figure above, the facsimile of an automobile is represented in basic shapes. This would make surface area measurements easier to make and calculate. Could your students estimate the surface area of an automobile?

Polynomial (Drafting: Architecture) Maximize area of a room In designing a museum, one of the galleries is originally to be maximized in area with a perimeter of 106 feet and be rectangular. This is for the display of art. Perimeter of a rectangle = 2x + 2y Solve for y 106’ = 2x + 2y 2 2 53’ = x + y 53’ – x = x + y -x 53’ – x = y continued Table of Contents Get the math teacher to supply you with similar problems.

Polynomial (Drafting: Architecture, cont.) Y, one side of the room, is equal to (53’ – x). Now use this Y in the Area formula: A = xy Substitute (53’ – x) for y. A = x(53’ – x) A = 53x – x 2 The unabbreviated form of the equation is: A = ax 2 + bx + c where a, b, and c are real numbers (coefficients) For our area equation in unabbreviated form, we get: A = -1x 2 + 53x + 0 So a = -1, b = 53, c = 0 This level of polynomial is always a parabola, mathematically. To find the maximum value (vertex) of a parabola use h = -b / 2a where a and b are the coefficients from our equation above. (It’s a downward facing parabola, based on the -1x 2. The h stands for height.) h = -53/2(-1) = -53/-2 = 26.5’ This will be used for x in the area equation A = 53(26.5’) – (26.5’ 2 ) = 702.25’ = maximized area h Table of Contents Get the math teacher to supply you with similar problems.

Trigonometry (Carpentry, Drafting: Architecture) Height of a Structure Introduce students to the basic trig functions. Use the right triangle pictured to help them get their bearings; what is a hypotenuse, opposite side, adjacent side… Explain that it is useful in architecture, transportation, and manufacturing when something has a triangle-related application and some information is not known. For example, the actual height of a building is not truly known. A surveyor can determine the height with the known information and an instrument. Could we use Tangent? We know the angle (θ), but for tangent, we have to know the opposite leg and the adjacent leg. Here we only have one unknown value, the opposite leg. By solving for the unknown (opposite) we could use tangent. Table of Contents

Simple Math, Area, Percent (Masonry) Load on a Scaffold Introduce students to the basic trig functions. Use the right triangle pictured to help them get their bearings; what is a hypotenuse, opposite side, adjacent side… A 180 pound mason is coursing brick on a massive chimney. She wants to keep about 100 pounds of mud on hand. Currently, there is only one level of scaffolding in use. On a 3’ x 8’ scaffold rated at 25 pounds per square-foot (load applied evenly across the scaffold), about what fraction of a 500 brick cube can the scaffold hold if the bricks weigh about 4 pounds each? Table of Contents

Simple Math, Area, Percent (Masonry, cont.) Area of the scaffold = 3’ x 8’ = 24 sqft. Total load supported = 24 sqft. X 25 lbs./sqft. = 600 lbs. Load already applied by mason and mud = 180 lbs. + 100 lbs. = 280 lbs. Load capacity remaining = 600 lbs. – 280 lbs. = 320 lbs. Number of brick allowed = 320 lbs. / 4 lbs. per brick = 80 bricks Percentage of a cube of about 500 brick = If 500 brick = 100 % Then 80 brick = ? Cross multiply and solve for ? 500? = 8000 500 500 ? = 16% Table of Contents Get the math teacher to supply you with similar problems.

Trigonometry (Masonry) Height of a Chimney Introduce students to the basic trig functions. Use the right triangle pictured to help them get their bearings; what is a hypotenuse, opposite side, adjacent side… Explain that it is useful in masonry, architecture, transportation, and manufacturing when something has a triangle-related application and some information is not known. For example, the actual height of a chimney is not truly known. The width is known but we can determine the height with the known information and an instrument. Could we use Tangent? We know the angle (θ), but for tangent, we have to know the opposite leg and the adjacent leg. Here we only have one unknown value, the opposite leg. By solving for the unknown (opposite) we could use tangent. Table of Contents