Examples of Math for Molders

1. What is the area of a square part that is 4 inches by 5 inches?

4″x5″ = 20 square inches

2. What is the area of a 1.5 inch diameter screw?

(Formula Diameter x Diameter x .7854= circular area)

1.5 x 1.5 x .7854 = 1.77 square inches

3. This same 1.5 inch screw mnning a 2.5″ shot size on a machine is producing how many Cubic Inches of shot size?

1.77 square inches x 2.5 inches = 4.43 Cubic Inches

4. Later you move the same tool above (4.43 Cubic Inch shot) to a machine with a 2″ diameter screw. What would the shot sized be to maintain the 4.43 cubic inches?

2″ x 2″ x .7854 = 3.14 square inches (screw area)

4.43 cubic inches / 3.14 square inches = 1.41 inches would be the equivalent shot size on the new machine.

5. The diameter of the injection cylinder is 7 inches, the diameter of the screw is 2 inches, what is the Intensification Ratio?

Ri=Ac/As where Ri: Intensification Ratio, Ac: Area of Cylinder, As: Area of Screw

7 x 7 x .7854 = 38.48 (Area of Cylinder)

2 x 2 x .7854 = 3.14 (Area of Screw)

38.14/3.14= 12.15:1 (Intensification Ratio)

Once you know the intensification ratio of a Hydraulic Machine, you can multiply it by any hydraulic pressure to calculate the plastic pressure (ppsi).

6. If the peak pressure at transfer is 18,060ppsi and we had a twenty percent rise in viscosity^ what would potentially our new peak pressure at transfer be?

18,000 xl.20= 21,600 ppsi

7. Our new mold is a two-cavity hot runner, the part is round (8 in diameter) with a .750 hole in the center. At three tons per square inch, how many tons are needed to support this mold minimally and we will be using a 400 ton machine?

8 x 8 x .7854 = 50.27 square inches minus .75 x .75 x .7854 (hole) = .44 square inches which equals 50.27-.44 = 49.83 sq inches x 2 cavities = a total area of 99.66 sq inches, x 3 tons per sq inch = 298.98 tons or 299 Tons

8. The toolmaker was asked to open the round gate from 0.025 in to 0.035 in. What was the effect on the area of the gate?

% of difference is the ((AREA of .025 – the AREA of .035)/AREA of .025) x 100 = OR

0.00049 – 0.000962/0.00049 = OR

0.000472/0.00049= 0.9632653 x 100 = 96.32653% difference

9. You are running a machine with a 1.25″ shot and the machines maximum injection stroke is 7.87″. What percentage of the shot are you using?

1.25″/7.87″ = .158831 x 100 = 15.8831% or 16% of the shot

10. You are running a 2.5″ shot at an injection speed of 3″/sec. How fast is it actually injecting if you f time is .93 seconds?

2.5″ / .93 = 2.68″/sec