how to use entitybox ?

Let's say you want to add an entitybox to a mesh of car...

I know you can use Meshdepth,Meshheight and Meshwidth to calculate the size of the mesh, but where do you start to calculate the entitybox?

Example:

the car mesh is 11,17,5 (width/height/depth)

Now, do I start from the center (5.5/8.5/2.5) or do I start
from 0/0/0 ?

And, if I use EntityBox, do I still have to use EntityRadius
of will EntityBox suffice?

Thanks

I don't mean to sound rude here, but have you even typed in these commands yet to see what happens. So many people come here asking questions before they've had a play at all.

It would take you 5 minutes of playing around with blitz to answer your own questions.

Why not give that a go?

Sorry for being blunt, it just annoys me.

If you had read the Collisions command you would have seen this:
So, the answer to that question is if you use EntityBox, you do not need to assign an EntityRadius (unless you want to).

Welcome back "Atomic Duck" - you haven't been around for a few months.

Keep at it "Atomic Duck", no harm in asking. I think "Rob Farley" and "Wolron" have been a little bit harsh (especially "Rob Farley" - no dessert for him tonight)

After all, it is 'Beginners Area'.

Here is a sample I had help with a while back and here is
a pretty good explanation I got.
http://www.blitzbasic.com/codearcs/codearcs.php?code=944
;1. Since you scaled the cube by 6 (the cube was originally from -1 To 1
;on it's Local axis) the -6's are the start position of the collision
;box (since 0,0,0 is the exact centre of the cube, in it's Local axis of
; course)...

;Not sure If I have explained that clearly enough - but essentially the
;midpoint of the cube is 0,0,0 - using EntityBox you have To specify the
; start point of the collision box (ie. one corner) And tell it the size.
; In this Case the New scaled up corner is now -6,-6,-6 And the New size
;is 12,12,12. It's much easier To grasp this If you draw it on paper - in
;effect it means that your collision box can be anywhere relative To
;your actual geometry.

;2. The collision normal will always be of length 1. Now, since my
; was moving by 0.2 units Each frame I just scaled down this
;length To make an arbitrary but reasonable effect of pushing the boxes
;at half the speed of the player - ie. 1/10 = 0.1 - this gives the effect
;that the boxes have friction against the Floor.

;In your instance you are moving the character by 0.25 units Each
;frame - therefore If you divide the collision normal values by 4 you
; will be moving the boxes at the exact speed of the player (1/4 = 0.25).

Although you may think my response was being harsh, it wasn't meant to be. I answered his question.

RiverRatt, WolRon

Thank you very much for your explanation.
The problem is pretty much solved now. :)