hehuvision: January 2012

Tuesday, 31 January 2012

How to calculate the distance between a arbitrary point to a plane?

refer to ( http://mathworld.wolfram.com/Point-PlaneDistance.html ) Point-Plane Distance

Given a plane

(1)

and a point

, the normal to the plane is given by

(2)

and a vector from the plane to the point is given by

(3)

Projecting

onto

gives the distance

from the point to the plane as

			(4)
			(5)
			(6)
			(7)
			(8)
			(9)

Dropping the absolute value signs gives the signed distance,

(10)

which is positive if

is on the same side of the plane as the normal vector

and negative if it is on the opposite side.

This can be expressed particularly conveniently for a plane specified in Hessian normal form by the simple equation

(11)

where

is the unit normal vector. Therefore, the distance of the plane from the origin is simply given by

(Gellert et al. 1989, p. 541).

Given three points

for

, 2, 3, compute the unit normal

(12)

Then the distance from a point

to the plane containing the three points is given by

(13)

where

is any of the three points. Expanding out the coordinates shows that

(14)

as it must since all points are in the same plane, although this is far from obvious based on the above vector equation.

Saturday, 28 January 2012

4 new NAOs are joining our group

Hi everyone,

We just got 4 new "guys" involve our group. Actually they are really good at acting performance, even dance better than me.

Here is the video demo made by my colleague Inkyu Sa.