Scratch

Makes sense. And yeah, amcerbu an you are the two people I'd probably trust most for Something like this XD. I'd suggest a way for it to inherent force from its "child" maybe?

amcerbu wrote:

Oh, yeah.  That stuff's really hard.  I saw a ragdoll physics article out on the internet awhile ago where "inverse kinematics" was concerned.  The way they explained it, kinematics is the study of determining the location of, say, the hand at the end of the arm, based on the angles of the joints.  Inverse kinematics tries to determine angles for the joints to produce a desired location for the hand.  I'm not exactly sure what that has to do with ragdolls, but you can check out the page for yourself. 

Anyway, here's the link.  Good luck!  Maybe I'll play around with something, but I'm not very hopeful that I'll get anywhere.

Thanks! Looks like a pretty good article.

Wes64 wrote:

You may find more help in Advanced Topics, even though this is the proper forum 

Maybe...but I'd rather try and figure it out with someone then just have someone tell me how to do absolutely everything.

Ok, so update on the idea: Position 1 for every part is calculated by the inheritance and what not.  Position 2 is then calculated with this:
Distance = sqrt( (1.x-2.x)^2 + (1.y-2.y)^2)
2.x = 1.x + ( (1.x-2.x)*length)/Distance
2.y = 1.y + ( (1.y-2.y)*length)/Distance
Where length is a predetermined number.  This avoids using trig and rotation which I feel will leave room for vector velocity with x and y components instead of a velocity based off direction and a magnitude.
It could also be used during the inheritance process as well.

Last edited by AtomicBawm3 (2012-07-12 13:34:52)

amcerbu wrote:

^^ I love vectors.  They make trig look so unnecessary.

Lol, yeah...still, trig does have to be used because I'm making the parts out of costumes...but it's minimal, only an arc-tangent really 

Are you talking about the stuff you posted above with distance constraints?  So far, I can't see anything wrong with it. 

A few other questions: what shape are you simulating?  A stickman?  Something more complicated?  Also, how fast do you want this to run? 

And here's an idea for implementation (at least, this is how I'd do it):  Everything is based on a central "source point."  For this project, the lines won't have mass, just the points, and we'll say all masses are equal.  Each point has an "impulse" velocity; the way it wants to travel.  It will try to travel that way.  When it reaches a distance constraint, it projects its velocity onto its constraints; any points it is attached to.  The remaining velocity (the vector "rejection") is given to the point at the other end of the constraint.  You may have to do some recursive stuff to get all the points to affect all the other ones.

If you manage to implement what the guy did in that article I posted, adding velocity won't be terribly difficult.

1s1sKing wrote:

Hmmm. What amcerbu said sounds about right. Just a suggestion, depending on how you want to do this, you could have it be one sprite an one script by stamping each body parts costume, or if it's simple, using the pen.

Yeah...I kind of started out that way, but I decided not to because it's easier sometimes to have calculations shown in separate broadcasts...and I'm running it very slow right now.  Still not working.  I'll upload it tomorrow to my test account and let you look.  It is a human body as well with 10 parts (upper and lower legs and arms, torso, and head).

Hey, AtomicBawm3: Check out this article. 

It has some interesting stuff about cloth dynamics and something close to ragdoll physics.  I hope you've had some Calculus in school (or, at least Pre-Calc) because there's stuff about Verlet Integration. 

I'd suggest you read the entire thing; it's pretty cool and you might be able use some of the things they talk about, especially where distance constraints are concerned. 

Gamasutra has some really good articles, so there's probably more out there.

Last edited by amcerbu (2012-07-13 00:19:58)

Ahhhh. Good idea. Another thing I would recommend you add is a little bit of force absorption from each "child". If say the lower arm gets flung, the lower arm should move. I hope I've been helpful 

@amcerbu I know that. I typically use that when I create a turbo mode project. It's also useful if you create a slider, and can has a user controlled framerate.

I'm checking out your "Pull Line" project right now.

I've gotta say, the rope with velocity looks really good.  Perhaps it would be nice if you included an option to adjust friction (or is it hard-coded into your script?).

Ok, I'm writing this because I'd like somebody to look over the math for this one final time...clearly I've already got most of the physics figured out, but I needed to figure out how to create offsets for the arms and legs so that they were a tad more realistic.  Here's the math I've come up with that avoids using trig to find a point rotated from any location:
Let's say that you've got a line for the middle of the body in the form y=mx+b based off of the two points at either end of the torso. Info needed:

Length of torso= 40 (for now)
m=(x1-x2)/(y1-y2) (of torso)
Offset on the y-axis (oy)= -4 (again, for now)
Offset on the x-axis (ox)= -10
and of course, x1, x2, y1, and y2

Basically, I'm going to calculate two vectors that are perpendicular to each other where the vector for y uses m and the vector for x uses -1/m.  Vector 1 can be calculated like so:
V1.X = -4(x1-x2)/40
V1.X = (x2-x1)/10

V1.Y = -4(y1-y2)/40
V1.Y = (y2-y1)/10

This vector is simple enough to calculate, but the perpendicular vector took some work:

First, you have to calculate a vector in the correct direction, which is simple enough since we know that a perpendicular line is simply -1/m.  Giving the vector the imaginary points of (x,y) and (x+1,y-1/m) we can clearly see that:
dx = -1
dy = 1/m
so our vector's current distance is sqrt(1/(m^2)+1).  To make the vector the right length, we simply do what I've been doing for all of the point calculations:
(x,y) of vector = (-dx*l/dist, -dy*l/dist)
so our vector becomes:
V2.X = -10/dist
V2.Y = -10/(m*dist)

To get the final position:
Final.X = Start.X + V1.X + V2.X
Final.Y = Start.Y + V1.Y + V2.Y
where Start is point 1.

Anyway...that's probably more confusing than it should be...but it should work.