Answer: Average shrinkage of plastic is 2% O Computation of relevant dimension of cavity Radial dimension:L=[Ln(1+k)-(3/4)△° =[10+0.02)-(3/4)×0.8]6 =1116+013 Depth Dimension: H=H1+k)-(2/3)△° [301+002)-(2/3)×0.3]3×16 304 Computation of relevant dimension of core Radial Dimension: 1=[, (1+k)+(3/4)Al-2 =[801+0.02)+(3/4)×06]06x 82.05 Depth Dimension: h=[h, (1+k)+(2/3)Als =[15(1+0.02)+(2/3)×0.2]025 =15430 Core Diameter: d=[d, (1+k)+(3/4)A] =[8(1+0.02)+(3/4)×0.1a:ss =8.240 3 Computation of position dimension of core C=Cn(1+k)±6/2 =30(+0.02)±(0.3×1/6)/2 =30.6±0.025 2. 1.3 Simplifying Method for Dimension Design Presently, almost all mold enterprises adopt three-dimension CAD/CAM to design mold, which turns out to be inconvenient. Therefore, they usually use simplifying method to calculateAnswer: Average shrinkage of plastic is 2% ① Computation of relevant dimension of cavity Radial Dimension: +δ L = [L（1+ k）− (3/ 4)Δ] p 0.8 1/ 6 [110(1 0.02) (3/ 4) 0.8] × = + − × 0.13 111.6+ = Depth Dimension: +δ H = [H（1+ k）− (2 / 3)Δ] p 0.3 1/ 6 [30(1 0.02) (2 / 3) 0.3] × = + − × 0.05 30.4+ = ② Computation of relevant dimension of core Radial Dimension: = + + Δ −δ l [l（1 k） (3/ 4) ] p 0.6 1/ 6 [80(1 0.02) (3/ 4) 0.6] = + + × − × 05 0.1 82. = − Depth Dimension: = + + Δ −δ h [h（1 k） (2 / 3) ] p 0.2 1/ 5 [15(1 0.02) (2 / 3) 0.2] = + + × − × 43 0.04 15. = − Core Diameter: = + + Δ −δ d [d（1 k）（3/ 4） ] p 0.1 1/ 5 [8 1 0.02 3/ 4 0.1] = （ + ）+（ ）× − × 24 0.02 8. = − ③ Computation of position dimension of core C = C（p 1+ k）± δ / 2 = 30(1+ 0.02) ± (0.3×1/ 6)/ 2 = 30.6 ± 0.025 2.1.3 Simplifying Method for Dimension Design Presently, almost all mold enterprises adopt three-dimension CAD/CAM to design mold, which turns out to be inconvenient. Therefore, they usually use simplifying method to calculate