Question

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Answer 1

GIVEN :- A 2D SKETCH ALONG WITH THE REQUIRED DIMENSIONS .

TO FIND :- THE OUTSIDE PERIMETER OF THE GIVEN GEOMETRY

SOLUTION:- The outside perimeter can be found out by the summation of length of small arcs in the periphery of the geometry and the straight lines included in the geometry . In order to make the calculations easier and understanding in a simpler manner , let the centers of the three holes be O1 , O2 and O3 , and let the perimeter be ABCDEQPA (refer image) .

Arc Ion +PA 30° 2S0 I.2 SU 0 radius 25 Bc- .963s

So the perimeter can be expressed as :

outside perimeter= AB+BC+CD+DE+EQ+QP+PA

The basic formula used in calculating the arc lengths is : arc= angle *radius of curvature , where angle is in radians . Using this formula the angle contained for arc AB is 75 degrees and radius given for arc AB is 7.250 hence the arc length of AB comes out to be = [75*pi /180 ]*7.250 =9.45

Similarly by using the same method , different arc lengths can be calculated as shown (refer image)

air 15ο DE = 0.3272 31 18o- 36.843-1 Nouw

Using the geometry and basic triangle properties , the required angles can be found out and thus the arc lengths can be found out .

NO- C om 90. 36.87. S3-13 2x s3-13。 lo 26 ļo PR is an aH Č.LWmnad, us 〈.s & angte jDE.26" L O' PQ463646 フ 3 1226 636464 3.1126 ANSw ER
And thus , the arc lengths of different parts come out to be as shown below :

AB= 9.45 , BC= 1.9635 , CD= 6.00 , DE= 0.32725 , EQ= 3.1226 , QP= 4.63646 , AND PA= 3.1226

On adding all these values , we get

outside perimeter = 9.45 + 1.9635 + 6.00 + 0.32725 + 3.1226 + 4.63646 + 3.1226

the required value of outside perimeter = 28.6224 ................................................................(ANSWER)