1. ## Perspective nightmare

Hi. I've been working in earnest through "Successfully Drawing", but recently hit a roadstop that's created some quite abit of confusing and stress.

-
I got to the chapter on shadows, and can do cubes no problem. But then it popped up, the sphere projection of the shadow is a tipped cube, with the sphere itself remaining on the ground plane.

I tried using an overlay on the upright axis, and tipping it in an attempt top make them all equal, and used the method earlier in the book with elevated roofs. After many hours, it's sort of a hodgepodge mess, I'm doing something wrong here and/or missing something, but I have no idea what it is.

Tried a good few times trying to tilt the cube up while keeping the sphere level, but it always turned into a distorting mess.

I need it tipped, so the axis remain correctly spaced, and the inside centre where the shadows shoots through to hit the surface also remains correct.

This has been driving my up the wall the last 2 days, I get of you sirs, please help me.

2. Hide this ad by registering as a member
3. What exactly do you mean by 'the sphere projection of the shadow is a tipped cube'?

4. The cast of the shadow hits the surface on the floor projected through the mid-point of the square/sphere.

But, as it hits the edge of the sphere (Loomis example) the angle of the cube needs tipped, but the sphere itself needs to remain on the surface.

The problem I am having, is tipping the cube, with it's axis in correct perspective so the sphere matches the projected light source, while keeping the sphere itself on the ground plane.

Basically, I need to tip the cube, while keeping the sphere in place. With the axis in perspective, without guessing.

I don't know how, I tried using a raised vanishing point but the entire thing was distorted.

5. Originally Posted by Riker Laforge
The cast of the shadow hits the surface on the floor projected through the mid-point of the square/sphere.
You lost me there.

6. I want to tip the cube while keeping the sphere inside it unmoved. That's all I need to know.

7. Ehh, so are you trying to make light fall mathematically right? Because, you know, that first picture right there, already has a major flaw in shadow casting logic.

Edit: Also, you don't need to add a new vanishing point for a tilted plane.
Last edited by White Rabbit; 4 Weeks Ago at 08:51 AM.

8. How do I do it? And what's wrong with it?

Not trying to be rude, but just saying "that's wrong", isn't very useful for someone who doesn't understand in the first place, very much a novice.

Help would be greatly appreciated.

9. It's just that I'm not quite sure that I get you. On one hand, you have the right idea, or at least that's what I assume, but that isn't reflected in your first picture. You'd have to align the lines for your light source with the corners of the tilted circle, not the sphere.
As for the tilting, you can use the existing vanishing points. Just draw a line and align 2 dots, do the same on a different height for the adjacent side and connect the dots.
Finally, I don't know how effective it is to spend this much time on shadows. I mean, sure it makes you understand this stuff better and all, but who has the time to do this for every drawing. Might as well pre-render in 3d, or just guess/ use reference.

10. ## The Following User Says Thank You to White Rabbit For This Useful Post:

11. Originally Posted by White Rabbit
It's just that I'm not quite sure that I get you. On one hand, you have the right idea, or at least that's what I assume, but that isn't reflected in your first picture. You'd have to allign the lines for your light source with the corners of the tilted circle, not the sphere.
As for the tilting, you can use the existing vanishing points. Just draw a line and align 2 dots, do the same on a different height for the adjacent side and connect the dots.
Thanks for the input. Still confused tbh, I'm trying to get it, but it's a mental brick wall.

When I tilted it in this picture, I used the mid point of ellipse within the the cube as fixed, so that when it tilted, the sphere within would tilt but not be raised.

I created a circle on the side plane of the sphere, and rotated the up/right round at an angle. And from that projected the lines to the VP points at different heights.

Upload the full picture as I'm pretty bad at explaining it, you'll prob see what I'm doing wrong.

I tried following this guys guide, and confused me even more.

12. Originally Posted by Riker Laforge
I want to tip the cube while keeping the sphere inside it unmoved. That's all I need to know.
You will need a bit more theory to construct that, specifically the use of measuring points. And, no, that is not what you need to know to get shadows more or less right. While I admire Loomis manuals, these are not good sources to study perspective. For anything practical, study 3D software, like blender.

13. ## The Following User Says Thank You to eezacque For This Useful Post:

14. Okay look, let's simplify that. All you need is a plane, the light source doesn't know if you have a sphere or not.
Last edited by White Rabbit; 4 Weeks Ago at 10:19 AM.

15. Again thanks.

But, how do I find the point where the light hits at the top accurately without guessing?

Also, would I be better just wholly dropping Loomis perspective book at this point and moving on to something more modern?

16. Make a tilted plane, by using the light source as a vanishing point, that'd make sense in my mind at least. Creating the tilted plane, based off the sphere's mid point.

17. Alright, thanks for the help chaps.

18. Just curious: why do you need the tilted cube?

19. To find the perspective of the sphere and light as it hit it's.

20. As you can see in your diagram, the terminator coincides only with the equator of the sphere if the light is infinitely far away, so in the case shown, your tilted cube is of no use...

21. It's not impossible but extremely impractical and time consuming (if you actually care about drawing) and you will need to understand how measuring points (MP) work, how to use them to measure slopes and lengths of foreshortened lines in space, and not just in 2PP but in 3PP. Check out books like Point of View - A Study in Perspective Drawing, Gwen White - Perspective: A Guide for Artists, Architects and Designers and https://www.handprint.com/HP/WCL/perspect4.html

For a source like the sun (approximated as being infinitely far away):
1) You will start with a cube with a sphere inscribed inside

2) Decide on the direction of the light source (along the horizon line), put a point there, draw a line from the point to the station point and draw a line perpendicular to this newly found line, intersecting it with the horizon line to obtain another point. This last point* will be one of vanishing point we will require later

3) Decide on the angle of the light source, use the measuring point for the vanishing point above which the light source should be (meaning each vanishing point has its own measuring point, and here we need MP for this particular VP) to plot the light source at the required angle (in the animation: intersection of pink dashed line at required angle from the horizon at MP(1,2) with the line perpendicular to the horizon through VP3/VP1)

4) Plot the line perpendicular to the line from MP to the light source (i.e. perpendicular to the pink dashed line in the animation), intersect this line with the line perpendicular to the horizon line through the light source (the line perpendicular to the horizon through VP3/VP1 in the animation) to obtain point**

5) Build a cube using 3 vanishing points: the point of light source, point* and point** with the same dimensions as of the original cube such that the centers of the two cubes coincide

6) Plot the intersection of the plane perpendicular to the light source with the cube through its center, which gives you a square in perspective

7) Use this square to plot a circle in perspective, this circle is the terminator (purple circle on the sphere in the animation)

8) Use it to find cast shadow

9) This second blue point below the blue center point of the sphere in the animation controls the position of the plane onto which the shadow is cast (so basically the height of the sphere above the ground plane, if we assume the first plane under the sphere is the ground plane)

Applet: https://www.geogebra.org/m/k9erzybg

Now all this can be done on paper, nothing besides knowledge of perspective was used.

22. ## The Following User Says Thank You to onemax For This Useful Post:

23. Originally Posted by onemax
It's not impossible but extremely impractical and time consuming (if you actually care about drawing) and you will need to understand how measuring points (MP) work, how to use them to measure slopes and lengths of foreshortened lines in space, and not just in 2PP but in 3PP. Check out books like Point of View - A Study in Perspective Drawing, Gwen White - Perspective: A Guide for Artists, Architects and Designers and https://www.handprint.com/HP/WCL/perspect4.html

For a source like the sun (approximated as being infinitely far away):
1) You will start with a cube with a sphere inscribed inside

2) Decide on the direction of the light source (along the horizon line), put a point there, draw a line from the point to the station point and draw a line perpendicular to this newly found line, intersecting it with the horizon line to obtain another point. This last point* will be one of vanishing point we will require later

3) Decide on the angle of the light source, use the measuring point for the vanishing point above which the light source should be (meaning each vanishing point has its own measuring point, and here we need MP for this particular VP) to plot the light source at the required angle (in the animation: intersection of pink dashed line at required angle from the horizon at MP(1,2) with the line perpendicular to the horizon through VP3/VP1)

4) Plot the line perpendicular to the line from MP to the light source (i.e. perpendicular to the pink dashed line in the animation), intersect this line with the line perpendicular to the horizon line through the light source (the line perpendicular to the horizon through VP3/VP1 in the animation) to obtain point**

5) Build a cube using 3 vanishing points: the point of light source, point* and point** with the same dimensions as of the original cube such that the centers of the two cubes coincide

6) Plot the intersection of the plane perpendicular to the light source with the cube through its center, which gives you a square in perspective

7) Use this square to plot a circle in perspective, this circle is the terminator (purple circle on the sphere in the animation)

8) Use it to find cast shadow

9) This second blue point below the blue center point of the sphere in the animation controls the position of the plane onto which the shadow is cast (so basically the height of the sphere above the ground plane, if we assume the first plane under the sphere is the ground plane)

Applet: https://www.geogebra.org/m/k9erzybg

Now all this can be done on paper, nothing besides knowledge of perspective was used.

Excellent. Exactly what was needed.

Thank you kindly sir.