Angular Perspective example :2 | Persepective Drawing

Hey Guys!! In the previous tutorial we basically talked about the perspective drawing in which the nearest vertical edge lied on the picture plane.
Now, let us take the following example in which the nearest vertical edge does not lie on the picture plane.
In this case, the cube of side 25 millimeters lies with its base on the ground and nearest vertical edge lying 10 mm above the picture plane.
Initially, draw a horizontal line representing picture plane.
Now, draw another horizontal line at an offset of 55 mm below the picture plane to represent horizon.
Also, draw another horizontal line at an offset of 40 mm below the horizon to represent ground plane.
Name the lines as picture plane, horizontal plane and ground plane as shown below.
Then, draw another horizontal line 10 mm above the picture plane and draw the top view of an object with a vertical edge on that plane as shown below. Now, name the vertices of the cube from 1 to 4 as shown below.
Since it is given that the stationary point lies in the central plane, 10 mm left to the center of the cube.
So, project the point 1 vertically downwards as shown below to intersect the horizon. Now, mark a point 10 mm left to the intersection point to obtain the required stationary point.
Now, simply join the stationary points with the edges 1, 2, 3 and 4 to obtain the intersection point a, b, c and d with the picture plane.
To obtain the vanishing points , simply draw lines from the stationary point, parallel to the line 1 2 and line 1 4 to obtain the intersection point with the picture plane.
Then, project the thus obtained intersection point vertically downwards to intersect the horizontal plane. This intersection points represents the two vanishing points.
Now, extend the line 1 2 up to the picture plane and mark that point as 5.
Then, project that point vertically downwards to intersect ground plane. This intersection point represents the point 5 on the ground plane.
Since the points 1 and 2 lies in the lines vanished from 5. So vanish the lines from point 5 as shown below; simply by joining point 5 with the vanishing points.
Now project the points a and b vertically downwards to intersect this vanished line to obtain the points 1 and 2 on the ground plane.
Then, since the point 4 lies in the line vanished from 1. So vanish the lines from point 1 as shown below.
Now project the point d vertically downwards to intersect this vanished line to obtain the point 4 on the ground plane.
Similarly, since the point 3 lies in the line vanished from either of points 2 or 4. So vanish the line from point 4 as shown below.
Now project the point c vertically downwards to intersect this vanished line to obtain the point 3 on the ground plane.
Let the points on the top face of the cube be represented as 1 prime, 2 prime, 3 prime and 4 prime.
Now, mark the point 5 on the top face of the cube; simply by drawing vertical line of 25 mm from point 5 as shown below.
Then, vanish the point 5 prime as shown below. And the intersection of the vanished line with the vertical line from a and b represents points 1 prime and 2 prime.
Proceed similarly to obtain the points 3 prime and 4 prime .
Finally, all the required vertices are obtained. Simply join all the vertices to obtain the required perspective drawing.

[@hazeno1thomas]

Thanks What if the picture plane is behind the plan?

[@AlkanBayraktar]

Hello can u try 3 dot perspective drawing in same technique?

[@rijankatwal1382]

How to make top hexagon

[@akankshasirsat4892]

Helpful! Thanks :)

[@Sandeep33]

😊☺