Cases Of Line In A.V.P And A.I.P & Profile Plane

Cases Of Line In A.V.P And A.I.P & Profile Plane

inclination of AIP with HP = inclination of FV with XY line

 

Line AB is in AIP as shown in above figure no 1. its FV  (a'b') is shown projected on Vp. (looking in the arrow direction)                                                                                                                                                                        Here one can see that the inclination of AIP with HP = inclination of FV with XY line

inclination of AVP with VP = inclination of TV with XY line

Line AB is in AVP as shown in above figure no  2...

Its TV (a b ) is shown projected on Hp. (Looking in arrow direction )

Here one can clearly see that the inclination of AVP with VP = inclination of TV with XY line

 

LINE IN A PROFILE PLANE ( MEANS IN A PLANE PERPENDICULAR TO BOTH HO & VP )

Results:-

1. TV & FV both are vertical, hence arrive on one single projector.


2. Its side View shows True Length ( TL )


3. Some of its inclinations with & VP equal to 90^. (θ  + Φ = 90^. )


4. it's HT& VT arrive on the same projector and can be easily located From side view.

OBSERVE ABOVE GIVEN ILLUSTRATION AND 2^nd SOLVED PROBLEM

PROBLEM 1:-

Line AB 80 mm long, makes 30^. angle with Hp End lines in an  Aux. Vertical plane 45^. inclined to VP. End A is 15 mm above HP and VT is 10 mm below XY line. Draw projections, fine angle with VP, and HT.

PROBLEM 2:

A line AB, 75 mm long, has one end A in VP. The other end B is 15 mm above HP and 50 mm in front of VP. Draw the projections of the line when the sum of its inclinations with HP & VP is 90^. , means it is lying in a profile plane. Find true angles with ref. planes and its traces. 

SOLUTION STEPS:

1. After drawing the XY line and one projector locate top view of A i.e point an on XY ass it is VP.


2. Locate FV of B I.e b' 15 mm above XY as it is above HP. and TV B i.e b, 50 mm below XY as it is 50 mm in front of VP Draw side view structure of VP and HP and locate S.V. of point B i.e. b''.


3. From this point cut 75 mm distance on VP and Mark a'' as A is in VP. (This is also VT of line.)


4. From this point draw locus to left & get a' Extend SV up to HP. it will be HT. Asit is a TV Rotate it and bring it on the projector of b.


5. Now as discussed earlier SV gives TL of line and at the same time on extension up to HP & Vp gives inclinations with those panes.

 

some cases of the line in different quadrants. ( try itself )

Remember:

1. below HP means FV below XY


2. behind VP means TV above XY.

1 ] T.V. of a 75 mm long line CD, measure 50 mm End C is 15 mm below HP and 50 mm in front of VP. End D is 15 mm in front of VP and it is above HP. Draw projection of CD and find angles with HP and Vp.

2 ] End of line AB is in HP 25 mm behind VP. End B in VP and 50 mm above HP. The distance between the projection is 70 mm. Draw projection and find its inclinations with Ht, Vt.

3 ] End A of a line AB is 25 mm below HP and 35 mm behind VP. Line is 300 inclined to HP. There is a point P on Ab contained by both HP & VP. Draw projections, find inclination with VP and traces.

4 ] End A of a line AB is 25 mm above Hp and end B is 55 mm behind VP. The distance between end projectors is 75 mm. If both its HT & VT coincide on XY in a point, 35 mm from the projector of A and within two projectors, Draw projections, find TL and angles and HT, VT.

Please write comments if you find anything incorrect and if you have any queries about this topic or you want to share more information about the topic discussed above.