# Problem 4.1 Projection of Points

## Problem 4.1 Projection of Points – Draw the projection of points, the position of as per data given below:

## (1)A point ‘P’ 25 mm above H.P. and 20 mm behind V.P.

## (2)A point ‘Q’ 20 mm below H.P. and 25 mm behind V.P.

## (3)A point ‘R’ 25 mm below H.P. and 20 mm in front of V.P.

## (4)A point ‘S’ 20 mm above H.P. and 25 mm in front of V.P.

## (5)A point ‘T’ on H.P. and 25 mm in front of V.P.

## (6)A point ‘U’ on H.P. and 25 mm behind V.P.

## (7)A point ‘V’ on V.P. and 20 mm above H.P.

## (8)A point ‘W’ on V.P. and 20 mm below H.P.

## (9)A point ‘X’ on H.P. as well as V.P. both.

**Procedure: **

**Procedure:**

* Step-1* Draw a horizontal line, which is x-y line of some suitable length.

* Step-2* Point P is 25 mm above H.P. and 20 mm behind V.P. So it is in 2

^{nd}quadrant and so on a single vertical line above & from x-y line, mark a point p’ at the distance 25 mm and point p at the distance 20 mm on it. Point p’ is front view & the point p is top view of the point P.

* Step-3* Point Q is 20 mm below H.P. and 25 mm behind V.P, so it is in 3

^{rd}quadrant. So draw a vertical line on both sides of the x-y line. On this vertical line, mark a point q’ 20 mm below x-y line and a point q 25 mm above x-y line. Point q’ is front view & the point q is top view of the point Q.

* Step-4* Point R is 25 mm below H.P. and 20 mm in front of V.P. So it is in the 4

^{th}quadrant. Draw a vertical line below & from x-y line. On this line mark a point r’ at the distance 25 mm and a point r is at the distance 20 mm from x-y line. Point r’ is front view & the point r is top view of the point R.

**Course:**Engineering Drawing

**Tutorial Type:**FREE

* Step-5* Point S is 20 mm above H.P. and 25 mm in front of V.P. So it is in 1

^{st}quadrant. Draw a vertical line on both sides of the x-y line. On this vertical line, mark a point s’ at the distance 20 mm above the x-y line & a point s at the distance 25 mm below the x-y line. Point s’ is front view & the point s is top view of the point S.

* Step-6* Point T is on H.P. and 25 mm in front of V.P. So it is in 1

^{st}quadrant but on H.P. Draw a vertical line below & from x-y line. Mark a point t’ on the point of intersection of the x-y line and the previously drawn vertical line & mark a point t on the vertical line at the distance 25 mm below x-y line. Point t’ is front view & point t is top view of the point T.

* Step-7* Point U is on H.P. and 25 mm behind V.P. So it is 2

^{nd}quadrant but on H.P. Draw a vertical line above & from x-y line. Mark a point u’ on the point of intersection of the x-y line and the previously drawn vertical line & mark a point u on the vertical line at the distance 25 mm above x-y line. Point u’ is front view & point u is top view of the point U.

* Step-8* Point V is on V.P. and 20 mm above H.P. So it is in 1

^{st}quadrant but on V.P. Draw a vertical line above & from x-y line. Mark a point v on the point of intersection of the x-y line and the previously drawn vertical line & mark a point v’ on the vertical line at the distance 20 mm above x-y line. Point v’ is front view & point v is top view of the point V.

* Step-9* Point W is on V.P. and 20 mm below H.P. So it is in 4

^{th}quadrant but on V.P. Draw a vertical line below & from x-y line. Mark a point w on the point of intersection of the x-y line and the previously drawn vertical line & mark a point w’ on the vertical line at the distance 20 mm below x-y line. Point w’ is front view & point w is top view of the point W.

* Step-10* Point X is on H.P. as well as V.P. both. So it is on the x-y line. Mark a point on the x-y line and give the names x’ & x on this same point. Point x’ is front view & point x is top view of the point X.

* Step-11* Give the dimensions by any one method of dimensions to all the points as shown into the figure.

In (8)W should be on XY and W ‘ should be 20 mm below XY

Good…Vishwa, Thanx.

It was just slip of click ..