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Axe / Re: Axe plasma
« on: May 04, 2012, 04:02:51 pm »
Yes, I did too. It's not hard. But that's not the point I'm trying to make
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Axe / Re: Axe plasma« on: May 04, 2012, 04:02:51 pm »
Yes, I did too. It's not hard. But that's not the point I'm trying to make
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Axe / Re: Axe plasma« on: May 04, 2012, 02:06:29 am »
I wonder if this concept could be helpful at all?
http://www.patrickmatte.com/stuff/physicsLiquid/ 33
Axe / Re: 256=0?« on: May 04, 2012, 01:15:07 am »
Basically he wants to only compare the LSB, as far as I can tell.
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Axe / Re: 256=0?« on: May 03, 2012, 11:50:44 pm »
I think he meant an 8bit comparison, not a 16 bit comparison...
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Math and Science / Re: Three intersecting parabolas« on: May 03, 2012, 10:59:27 pm »
If the foci are all at the same ycoordinate, they will never intersect no matter what the value of d is. As long as their ycoordinates are not all the same, it is solvable.
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Math and Science / Re: Three intersecting parabolas« on: May 03, 2012, 10:17:40 pm »
Yes, they all have the same directrix. The foci can all lie on the same line, as long as that line is not parallel to the directrix.
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Math and Science / Three intersecting parabolas« on: May 03, 2012, 08:01:19 pm »
I've been thinking about this all day and I can't figure it out...
Say you have three known points p_{0}, p_{1}, and p_{2}. Each of these points serves as the focus of a parabola, and all three parabolas share a single line at y=d as their directrix. Assume that d is such that all three parabolas open in the same direction, and assume that the three foci do NOT all lie on a single line parallel to the directrix. So the idea is to solve for value of d that makes all three parabolas intersect at a single point. There should only be one value that makes this true. Basically, so far I've got quite a bit conceptually figured out in terms of what needs to happen...but I don't really know how to make it happen. Here's kinda what I have: The intersection point(s) of two parabolas (z_{n})can be calculated fairly easily: z_{0}=a_{0}x^{2}+b_{0}x+c_{0} z_{1}=a_{1}x^{2}+b_{1}x+c_{1} Find the quadratic equation that is the difference of the two, and set it equal to zero: 0=(a_{0}a_{1})x^{2}+(b_{0}b_{1})x+(c_{0}c_{1}) Use the quadratic formula to find the zeros: x=((b_{0}b_{1})±√((b_{0}b_{1})^{2}4(a_{0}a_{1})(c_{0}c_{1})))/2(a_{0}a_{1}) A parabola can also be written as (xh)^{2}=4p(yk), where (h,k) is the vertex of the parabola, (h,k+p) is the focus, and the directrix lies at y=kp With a focal point F and a directrix at y=d, all other necessary variables can be calculated as follows: p=(F_{y}d)/2 h=F_{x} k=F_{y}p a=1/4p b=h/2p c=h^{2}/4p+k Through this, all variables can be simplified down to only use F and d: a=1/2(F_{y}d) b=F_{x}/(F_{y}d) c=F_{x}^{2}/2(F_{y}d)+F_{y}(F_{y}d)/2[/tt] So you could technically write out the equation of a parabola as y = x^{2}/2(F_{y}d) + F_{x}x/(F_{y}d) + F_{x}^{2}/2(F_{y}d)+F_{y}(F_{y}d)/2 As you can see, it gets pretty hectic pretty quickly. For example, the quadratic formula would then be: x=(F_{x}/(F_{y}d)±√((F_{x}/(F_{y}d))^{2}(2/(F_{y}d))(F_{x}^{2}/2(F_{y}d)+F_{y}(F_{y}d)/2)))/(1/(F_{y}d)) Remember, this needs to be solving for d... Really, I don't know where to go from here. Anyone have any insight? EDIT: Okay, so I'm slowly figuring things out. The equation is crazy complex, but you have to solve for the intercepts of two parabolas, then plug that quadratic formula calculation in for X in one of those equations, and set it equal to the equation for the third parabola with the quadratic formula calculation plugged in for X as well. a_{2}(((b_{0}b_{1})±√((b_{0}b_{1})^{2}4(a_{0}a_{1})(c_{0}c_{1})))/2(a_{0}a_{1}))^{2}+b_{2}(((b_{0}b_{1})±√((b_{0}b_{1})^{2}4(a_{0}a_{1})(c_{0}c_{1})))/2(a_{0}a_{1}))+c_{2}=a_{0}(((b_{0}b_{1})±√((b_{0}b_{1})^{2}4(a_{0}a_{1})(c_{0}c_{1})))/2(a_{0}a_{1}))^{2}+b_{0}(((b_{0}b_{1})±√((b_{0}b_{1})^{2}4(a_{0}a_{1})(c_{0}c_{1})))/2(a_{0}a_{1}))+c_{0} Looks like fun, right? Now substitute the As, Bs, and Cs with their F/d equivalents that I mentioned earlier, accounting for the fact that the Fs have to be F_{nx} and F_{ny} for each variable of the corresponding n value. 38
News / Re: Some site features temporarily disabled« on: April 27, 2012, 05:02:58 pm »
Lack of search makes things really difficult when there's a specific thread that you want to find...Is there really no way to implement it back in?
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Art / Re: General purpose art thread« on: April 27, 2012, 06:00:56 am »
Here's one I did a while ago.
Everything's done with only red, blue, yellow, and black. I do use both a dark blue and a light blue, however. If for some reason you want to see it, here's a larger version. 40
TI Z80 / Re: Dev Team for Zelda« on: April 27, 2012, 01:25:17 am »apart from bosses, the ai really isn't complex. enemies wander around, turning at random, and shoot/attack at random iff you are in front of them.Exactly...but the bosses are gonna be a pain... 41
TI Z80 / Re: Dev Team for Zelda« on: April 26, 2012, 09:31:17 pm »
Well what about the AI? It is certainly more complex than just constantly moving toward the player. Sure, some enemies do that, but there are certainly more complex actions as well.
What enemies are going to be included in the game? That will help to figure out what needs to be done for the AI. 42
TI Z80 / Re: Dev Team for Zelda« on: April 26, 2012, 03:33:00 pm »
Bullets for hookshot and bow would only be 4 direction, Xeda.
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Math and Science / Re: Polygon algorithm« on: April 15, 2012, 11:14:45 pm »
Triangulate, then remove unnecessary vertices. That's honestly the best way to do it with any simplicity. Anything more and you'll be shooting yourself in the head.
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Humour and Jokes / Re: scary picture« on: April 12, 2012, 08:51:18 am »wut? now i'm totally confused Ö you never see the camera turn, ans still it must have turned around..dude my mond has been blown now, totally Don't look if you don't want to know how it was done... Spoiler For how they did it: 
