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c’1
                                                                                                                b’1
   Problem 8: A circle of 50 mm diameter is                a’        b’ d’       c’
   resting on Hp on end A of it’s diameter AC                                             300                  a’1           d’1 Y
                                                  X
   which is 300 inclined to Hp while it’s Tv                                                   d1
                                                                                                           450
                                                                     d
   is 450 inclined to Vp.Draw it’s projections.

                                                           a                      ca                     c1
                                                                                      1

Read problem and answer following questions
1. Surface inclined to which plane? -------      HP                     b                      b1
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV              The difference in these two problems is in step 3 only.
4. Which diameter horizontal? ----------         AC        In problem no.8 inclination of Tv of that AC is
   Hence begin with TV,draw rhombus below                  given,It could be drawn directly as shown in 3rd step.
   X-Y line, taking longer diagonal // to X-Y              While in no.9 angle of AC itself i.e. it’s TL, is
                                                           given. Hence here angle of TL is taken,locus of c1
                                                           Is drawn and then LTV I.e. a1 c1 is marked and
                                                           final TV was completed.Study illustration carefully.
  Problem 9: A circle of 50 mm diameter is
  resting on Hp on end A of it’s diameter AC
  which is 300 inclined to Hp while it makes                                                                           c’1
  450 inclined to Vp. Draw it’s projections.                                                             b’1
                                                      a’        b’ d’       c’
                                                                                                     a’1              d’1
                                                                d                         d1
                                                                                                               300
       Note the difference in
                                                      a                      ca                     c1
      construction of 3rd step                                                   1

         in both solutions.
                                                                 b                        b1

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Projection of planes 010

  • 1. c’1 b’1 Problem 8: A circle of 50 mm diameter is a’ b’ d’ c’ resting on Hp on end A of it’s diameter AC 300 a’1 d’1 Y X which is 300 inclined to Hp while it’s Tv d1 450 d is 450 inclined to Vp.Draw it’s projections. a ca c1 1 Read problem and answer following questions 1. Surface inclined to which plane? ------- HP b b1 2. Assumption for initial position? ------ // to HP 3. So which view will show True shape? --- TV The difference in these two problems is in step 3 only. 4. Which diameter horizontal? ---------- AC In problem no.8 inclination of Tv of that AC is Hence begin with TV,draw rhombus below given,It could be drawn directly as shown in 3rd step. X-Y line, taking longer diagonal // to X-Y While in no.9 angle of AC itself i.e. it’s TL, is given. Hence here angle of TL is taken,locus of c1 Is drawn and then LTV I.e. a1 c1 is marked and final TV was completed.Study illustration carefully. Problem 9: A circle of 50 mm diameter is resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it makes c’1 450 inclined to Vp. Draw it’s projections. b’1 a’ b’ d’ c’ a’1 d’1 d d1 300 Note the difference in a ca c1 construction of 3rd step 1 in both solutions. b b1