2018 RETA Breeze May- June

Continued from page 13

packages would have a dimension of 24” wide (3x8), 14” long (1x14) and 3” high (1x3). ree packages per row. Now I multiply the height 3” x the number of rows needed 8 (24 presents / 3 per layer) = 24”. So, my presents form a stack 24” x 14” x 24”. e depth of the closet is 27” (deeper than what we need (24”)) by 18” wide (wider than what we need (14”)) and 48” tall (taller than our stack of presents (24”). SO our closet would hold twice as many presents as we need. Double presents? 1.5 We calculate the volume of the room. 60’ x 35’ x 20’ = 42,000 cubic feet. 1.6 e diameter of the pulley 14” x 3.1416 (Pie) = 43.97 inches. 1.7 e circumference is 4.7124” What is the OD of the tube? We divide the circumference by pie (3.1416) 4.7124 / 3.1416 = 1.5” e diameter of the tube. 1.8 e circumference of the pipe at 20.81”. to nd the diameter we divide the circumference 20.81 / 3.1416 (Pie) = 6.625 Looking at the chart we determine it is 6” pipe.

Volume calculations of tubes can be very useful. Volume of tubes are similar to the volumes of boxes. We multiply the area by the length. For instance, a tube 4” in diameter and 5’ feet long would have a volume of 753.9 inches or .136 cubic feet. 1.10 We are ordering concrete for 30 bollards. Each bollard utilizes 8” Schedule 40 pipe. e pipes will be 6’ long. What is the total volume of the pipes? How much concrete should we order? If you are still having trouble with these calculations, a website https://www.ixl.com/math/ has hundreds of math problems available and you can pick the types of problems you need help with. Next time, we’ll look at rates. ANSWERS: 1.1 e door is 6’ wide, the wrapping machine is 8’ wide. e hole (doorway) (6’) – the object (8’), 6-8 = -2, the door isn’t wide enough. e time to nd this out is before the wrapping machine arrives. 1.2 To paint a room 130’ long x 35’ wide. We must calculate the square footage. 130 x 35 = 4,550 sq. . en we divide the number of square feet of oor (4550) by the amount that can be painted with a gallon of paint (1500). 4550/1500 = 3.03 gallons. So, we need to buy 3 gallons and an extra quart. 1.3 We need to store 13 pallets (4’ x 4’) in a room, the pallets cannot be stacked. e room is 22’ wide and 31’ long, can the pallets t? ere are several ways to do this. is is the method I would use. How many pallets can I put in each row? 22’ wide / 4’ = 5.5 We can put ve pallets in a row (we can’t cut the pallet). How many pallets can I put in each column? 31’ / 4’ = 7.75 We can put seven pallets in a row (we can’t cut the pallet). 1.4 Will 24 presents (8” wide, 14” long, and 3” high), t in a coat closet (27” deep, 18” wide and 4’ tall)? ere are a couple of ways to do this problem as well. I like to stack my presents neat and I look at how many presents I can get per row. I can put 3 presents per row. 3 x 8” = 24” So three

1.9 What size hole do we drill to accommodate a new 1” Schedule 40 A53 water pipe?

is is a straight lookup question; no math is involved. But we still need to think about it. e chart says we need a 1.315” Dia. Hole saw. But since this is the engine room wall we will need room to re caulk it. I would select a 1.5” hole saw. If I were to install a wall sleeve, we need to put even more thought into it. To accommodate my 1.315” outer diameter water pipe I would need a pipe for the wall sleeve with at least a 1.315” inner diameter. Looking at our chart I would select a 1-1/2” Pipe for the wall sleeve. Now to accommodate the 1.900” outside diameter of my 1-1/2” pipe I would need a 2” hole saw.

14 RETA.com