### Qustion about 3D

Posted:

**Sun Apr 16, 2017 4:55 am UTC**What is the best way to represent 3D in 2D?

There is a lot of ways to daw 3D in 2D and which one is the best?

There is a lot of ways to daw 3D in 2D and which one is the best?

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Posted: **Sun Apr 16, 2017 4:55 am UTC**

What is the best way to represent 3D in 2D?

There is a lot of ways to daw 3D in 2D and which one is the best?

There is a lot of ways to daw 3D in 2D and which one is the best?

Posted: **Sun Apr 16, 2017 8:15 am UTC**

Tevin Chung from South Korea wrote:What is the best way to represent 3D in 2D?

There is a lot of ways to daw 3D in 2D and which one is the best?

(Decided to include quote, as this might end up being transferred to a "loads if little questions" thread (in coding), as with the ATP one kin Science), etc.. )

What are you wanting to represent? One/two/three-point perspectives have their own interesting features, isometric projection can be technically good, other obliques have their own uses, whilst front/top/side views (in US or UK formats, i.e. whether you 'rotate' the object above or below the paper) with the fourth quarter an optional perspective/isometric/oblique overview (or indeed any free-camera version, with the same decision tree as before) is good for conveying maximum information, at the expense of making the space for the details smaller.

But, with computing power behind the visualisation (without having to work it all out with set-square and rulers), camera-style perspective can be done with a 3x3 or 4x4 matrix transform of (modelled) x, y, z onto (displayed) x, y (and optional depth-of-field effects that one might want to apply).

Posted: **Mon Apr 17, 2017 5:35 pm UTC**

(x' = y' = 0 is trivial, so I'll assume it's not considered)

x' = tanh(x+1)*y*z+1.37+(x+z)*ln(1+x^2+y^2+z^2)

y' = tanh(x)*z-cos(z+x)*y

I dare you to give a better projection.

x' = tanh(x+1)*y*z+1.37+(x+z)*ln(1+x^2+y^2+z^2)

y' = tanh(x)*z-cos(z+x)*y

I dare you to give a better projection.

Posted: **Mon Apr 17, 2017 9:15 pm UTC**

Flumble wrote:(x' = y' = 0 is trivial, so I'll assume it's not considered)

x' = tanh(x+1)*y*z+1.37+(x+z)*ln(1+x^2+y^2+z^2)

y' = tanh(x)*z-cos(z+x)*y

I dare you to give a better projection.

Please show how this projects the Utah teapot