Recent content by brimby

Yes.

Noted. For now I'll probably just cheat and set Δx and Δy equal to some constant. I'll work on the block tipping properties next. I've noticed there are some good discussions already laid out on that at this forum.

But don't worry, once I finally get through this initial exercise, I'm going to give the rectangle (and the people of this forum) a break and work on some other stuff for a while.

Yeah, that's my problem. I was saying "the rotationVelocity (Δθ) of the object is equal to the attackVelocity of the player" but there is no reason why that would directly translate. I need to make the attackVelocity affect the x and y velocity of one side of the rectangle, and calculate the θ...

You know what, scratch that whole last ramble. I've decided that switching the directly affected values to Δcos and Δsin is definitely the way to go. It doesn't make sense that the character hits the rectangle and its degrees of rotation is what is directly affected. The forward motion of the...

Ok I think I have the φ issue squared away. I think I really just have one last hurdle here, and it has to do with making things easier for programming. Generally I am working with Δvalue because it's way easier for me to program things to change an amount from where they're at, rather than...

No I think we're doing the same. I'm just considering \varphi to be 0 and therefore θ = 0. I'm just setting the axis to go through the center diagonal of the rectangle. Is that okay? Hmmm... maybe not. Cuz then it wouldn't be sitting flat at the end of the turn, and that would cause further...

O ok! In the first corner, Δx is positive as x shrinks from r to 0 and as θ increases from 0 to π/2, and in the second corner Δx is positive as x grows from 0 to r and as θ shrinks from π/2 - 0 to π/2 - π/2! I'll have to think about the implications of this on my formula... And the reason I...

I understand your second point clearly. Your first one I'm still pondering. Are you trying to say that I should do π - Θ rather than π/2 - Θ for the "evens"?

Ok great! And I'm going out on a limb right now because I feel like I might be on a roll (wait, am I on a limb or a roll??), but for the second quarter turn is it: Δx = r(Δcos(π/2-θ)) Δy = r(Δsin(π/2-θ)) ??

O jeez. I didn't realize there is a calculus component that I hadn't even made it to yet. So you're telling me a that it's much more complex than just: Δx = r(Δcosθ) Δy = r(Δsinθ) for the first quarter turn??

It only seemed that way to me because in that specific case where x1 is the very first point in the turn the change in r(1-cosΘ) is the same as the change in rcosΘ because x1 = r. Or something like that. Oops.

Yeah I thought about it for 2 seconds and you're right. Change in x equals the change in cosΘ. I'm a goof. Thanks though.

∆x is the x position of the center of the rectangle minus the previous x position of the center of the rectangle. My rectangle isn't going to have the luxury of sitting at (0,0) in my game. It could be anywhere on the coordinate system, which is why I opt to use ∆x so I don't have to figure out...

Yeah I definitely don't want my coordinates to have an origin at the edge of the circle. But wouldn't my version work in terms of Δx? My example rectangle is lying horizontally at the start, and rolling to the right. So that x at the edge of the circle would be a qualifying Δx as it moves right...