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# Ecoste

Member Since 22 Sep 2011
Offline Last Active Jan 28 2015 11:07 AM

### #5190494Affordable mobile game (iOS/Android) marketing strategy/company

Posted by on 31 October 2014 - 06:16 PM

I've always wondered about advertising mobile apps. The only ads I ever see for them is on other apps or some crappy site pop-ups. Maybe on FB as well, but I've never had a Facebook account, so I have no idea. I mean, you can't expect many downloads if you just put it on the app store and leave it be.

Are there companies that specialize in this kind of promotion?

### #5090769Designing levels...

Posted by on 01 September 2013 - 06:57 AM

Hello, I had an idea for a flash game, basically it's a puzzle game that's tile-based. Each level has a finish that you must get to, but to get to it you need to place tiles that are given to you. I have a prototype going and it's not really a code intensive game, but the problem is creating the puzzles. Like, I only really thought of 1 or 2 levels but I can't come up with anything anymore, how do you do it? I mean, you could always "Borrow" them from other games and then take out some tiles to place, this is the last resort though.

### #4954873Extending a normalized vector?(3d)

Posted by on 02 July 2012 - 05:36 AM

Ok, again, where C is the camera position, and N is the unit look-at vector...

It's the same thing in 3D. If you want your vector to be 2 or 3 times longer, then multiply all of the vector's components by 2 or 3.

To draw your unit length red line:
Start at C and end at C + N.

To draw a longer line that includes the blue:
Start at C and end at C + N*3.

As for determining 2 or 3, or whatever the factor is... that's application dependent, and it's really your call. Best of luck.

I notice that the function call drawLine(fromAllCharacters, vector3); basically treats vector3 as a position, when it's a displacement (ie. a direction times a length). Not sure what you're trying to achieve here, since it isn't at all like what I tried to show you in the first two examples. Note how I end the line segment at C + N*d (ie. a position plus a displacement is a position), not at N*d (ie. a displacement plus nothing is a displacement). These line segment drawing functions deal in positions, not displacements.

A vector V = <V.x, V.y, V.z>, or displacement, is essentially a 1D line segment (or arrow, if you prefer) that always starts at the origin's position O = <0, 0, 0> and extends to the position <O.x + V.x, O.y + V.y, O.z + V.z>. The vector is an extended object.

A position P = <P.x, P.y, P.z> is essentially a 0D point that "starts and ends" at <P.x, P.y, P.z>. The position is not an extended object.

Both objects are defined using three components, and you can even make these components equal in value by setting P = V, but in the end they are not the same kind of object. A displacement is an extended object, a position is not. The start and end points of a line segment are not extended objects.

Thanks a lot, it worked. Sorry for not understanding you earlier :/

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