Metroid Pi

[Harrierfalcon]

Urghh......*foams at mouth, takes swig from flask*

*hiccups*

I'm kinda busy right now, if I get time I'll release what I have so far...I might rewrite the second boss, he's really lame right now.

[IAmACalculator]

For all your motivational needs, go to despair.com!

[Harrierfalcon]

Ha ha...engineers smell bad...tee hee

j/k :D

I am motivated now, though. I'm rewriting the second boss to something a little cooler than a theta that follows you around.

[frenchcalc1]

[Harrierfalcon]

Thanks! The second boss is a Dark Samus-kind of boss--he follows you an shoots at you. I'm having a few problems with the shooting part, but I'll figure it out.

[garrett01001]

for the last boss you should do something similar to the following:

have a weird looking sprite, which is really just pi and e overlapped, then have them separate and pi shoots 3.141592etc (pi) at you and e shoots 2.71828etc (e, eulers number) at you

just an idea.. feel free to change it if you want

[Harrierfalcon]

Well, it has to be a plain Pi symbol, because overlapping symbols would take too long, and the strongest enemy is a Pi (hence, Metroid Pi)...

Ok, I am stuck. I want the theta to shoot a beam of stuff at the character (Samus?). The theta follows the same gravity and jumping power Samus does. How would I come up with a formula that would give me the length from the theta to the wall, (The walls are at 0 and 11), factoring in which way Samus is in regards to the theta?

[garrett01001]

theta position = 7
samus position = 3
shoot = 7,6,5,4,3,2,1,0 (stops moving when it hits something)

[Harrierfalcon]

Uhhh....that's not exactly what I meant. I wanted a formula to give me the length from the theta to the wall. Not for the shot to move, just a beam.

[Weregoose]

If there are two axes, find the hypotenuse. If there is one axis, simply subtract the wall position from the theta position and take the absolute value. If you want the wall in the direction away from Samus, then it'd make sense that if θ_pos minus S_pos is positive, then it'd be one way; if it's negative, then it'd be in the other direction. Make sense?

[garrett01001]

11-(current pos) = distance to right side
(current pos)-1 = distance to left side

?

[Harrierfalcon]

Umm....OK, let me clarify.

Say I had homescreen-sized graphics, but with the top left corner being (0,0) instead of (1,1). The theta is at (6,5) (row,col). How would I get the distance from the theta to (6,1) or (6,10)?

EDIT: Hrmm....I was attempting to avoid any If-Thens....I'll make this work. Thanks guys!

[garrett01001]

use basic distance formula:

sqrt((x2-x1)^2+(y2-y1)^2)

(i think thats right..)

[Harrierfalcon]

Umm....sure? That's a too complicated.

I think with a clever combination of Booleans I can make this work. Thanks!

[garrett01001]

complicated... that's super easy..

you are at (6,2) you want distance to (5,3)

sqrt((5-6)^2+(3-2)^2)
= sqrt(1+1)
= sqrt(2)

can't get much better than that?


EDIT: and of course, if the X or Y is the same then you just subtract the two that are different..

[Xphoenix]

I've been unable to shoot, so I was wondering if anyone else has experienced this?

[Weregoose]

Assuming I have this right:

Define [font="courier new;font-size:9pt;line-height:100%;color:darkblue"]θ

[Harrierfalcon]

Yay! Thanks guys!

@Xphoenix: That is part of the game. You have to get the powerup that lets you shoot.

[Xphoenix]

Oh. So you can change screens... I must be a slow learner.
(That's what readmes are for)

[Harrierfalcon]

........:|

Anyways...I've finished the second boss, and I've decided to update the demo. Should be self explanatory, controls are simple...any suggestions?

Here:[attachment=1455:attachment]