Physics question

[BasicCoder2]

fig. 4 and fig. 5 and fig. 6 are taken from my latest program where I am using a single rocket motor that can be rotated.
fig. 6 I would suggest is equivalent to your first example where you have two fixed motors with one not firing.
I would suggest that the vector force would be divided up as shown in fig. 3

[deltarho[1859]]

With sin(0) = 0, cos(0) = 1, sin(90) = 1 and cos(90) = 0

From second post with beta = 0 we have

towards centre = F
angular momentum = 0

This corresponds to BasicCoder2's fig. 5

With beta = 90 we have

towards centre = 0
angular momentum = F

This corresponds to BsicCoder2's fig. 4

fig. 3 corresponds with the second post.

I cannot see what BasicCoder2's last post is adding more to the conversation than the second post did; or am I missing something?

[badidea]

I cannot see what BasicCoder2's last post is adding more to the conversation than the second post did; or am I missing something?

Neither did I.

As I don't see a real difference between the flying saucer and the borg cube, I expect both to move in y-direction only while rotating. (assuming the thruster fires shortly, like impulse). Working on some code now...

Actually I build a test setup: A piece of wood (40 cm long), hanging on a thin flexible wire (2 m long), with the wire connected to the center so the the piece of wood is exactly balanced. Then I hit it with an wooden spoon firmly, but short "tak". Although difficult to perform accurately, the suspended piece of wood seems to start swinging only in the direction of the wooden spoon movement. Plus rotation.

[BasicCoder2]

I cannot see what BasicCoder2's last post is adding more to the conversation than the second post did; or am I missing something?

Neither did I.

As I don't see a real difference between the flying saucer and the borg cube, I expect both to move in y-direction only while rotating. (assuming the thruster fires shortly, like impulse). Working on some code now...

Some nice pictures to reference :) I tend to think visually.

Last night I was watching a documentary about Michael Faraday and how his lack of math limited his genius. Instead it was Maxwell that wrote the famous equations after reading the works of Faraday.

What I was hoping to do next is instead of having keys for linear and angular velocity control of separate rocket engines (which I think they use to work the lunar lander mother module) have keys that swivel the rocket engine and then fire an impulse from the rocket engine with a tap of the space bar to see how it looks. As I alluded to before, like outboard motors on a speed boat.

The borg cube is of course 3d and a lot more complicated. Hitting a circle is the same at any point. The angle you hit a square will depend on the orientation of the square. A frictional contact is also required I think to add angular momentum otherwise the force is entirely through the centre of mass. An attached rocket engine has 100% frictional contact.

Ideally someone who has done a course in math and physics should be able to write the code that can be used in an actual game.

[badidea]

The borg cube is of course 3d and a lot more complicated.

You are allowed to squeeze to borg cube to a height of nearly nothing :-)

Hitting a circle is the same at any point.

I am not hitting the circle. I have attached engine which shortly applies thrust in y-direction.

The angle you hit a square will depend on the orientation of the square.

I hit the cube perpendicular. Haven't decided between firing a billiard ball at it or a picee of sticky gum.

[badidea]

Here is a general strike.
I have used the property of mass of a disc and moment of inertia of a disc.
...

Looking at you code, but I is a bit hard to follow. Not written for a beauty contest :-)

Code: Select all

linearimpulse = impulse * Cos(a) / mass
angularimpulse = impulse * Sin(a) / anginertia

You assume that linear impulse is in cos(a) direction and angular impulse in sin(a) direction. But that is were a have my doubts, unless simulation pool billiard.

[D.J.Peters]

(The laws of physics apply everywhere), maybe not here.

May be force are not with you if you don't understand :lol:

Again the cross product (d,F) was and is the key

d=direction from center to point of interaction

F = the force acting on the point

In 3D space the result of the cross product is an vector and the most important property of this vector are the magnitude.
if
Speed = magnitude(velocity)/timestep
then
AngularVelocity = magnitude( cross(d, F) ) ' in radiant per second
here from my code:

Code: Select all

 ...
' german: Winkelgeschwindigkeit 
type tAngularVelocity  as tReal ' = cross(d,F)
' german: Drehmoment
' Torque is a measure of how much a force acting on an object causes that object to rotate.
type tTorque           as tVector ' = angular acceleration
...

The tricky part is only how to get the cross product in 2D space ?
here from my 2d vector class

Code: Select all

' pseudo cross product last row from 3D cross product !
operator \(l as tVector, r as tVector) as tReal
  return l.x*r.y - l.y*r.x
end operator

AngularVelocity = magnitude(d \ F) ' "\" overloaded operator for cross product

Joshy

Do you know ?
from: Video Game Physics Tutorial - Part I: An Introduction to Rigid Body Dynamics

[BasicCoder2]

I expect both to move in y-direction only while rotating.

You may be right with an object hitting a Borg ship. The boat and it's motor interacts with the water it is not skidding along a frictionless frozen lake with a rocket for propulsion.
https://www.youtube.com/watch?v=qaIqanZ00uo
However if there was a rocket engine on the block it would turn with the block and thus change the direction of the force in absolute space?
So to test this you would need to have a rocket of some kind (water bottle with vinegar and baking soda dissolved in water?) and a low friction setup such as a disc on an ice rink.

[dodicat]

Here is a general strike.
I have used the property of mass of a disc and moment of inertia of a disc.
...

Looking at you code, but I is a bit hard to follow. Not written for a beauty contest :-)

Code: Select all

linearimpulse = impulse * Cos(a) / mass
angularimpulse = impulse * Sin(a) / anginertia

You assume that linear impulse is in cos(a) direction and angular impulse in sin(a) direction. But that is were a have my doubts, unless simulation pool billiard.

When the angle is zero all goes into the linear, so cos.
When the angle is 90 all goes into the turning so sin.
To use a cross product in I just add a .z component.
I only use the vector products here to deduce an angle in the range 0 to 360.
The cross product uses a bit more cpu, but with the cross product the dot product is not really needed to return an angle.
Angle example:

Code: Select all

  Type pt
    As Single x,y,z
End Type

Function length(p1 As pt) As Single
    Return Sqr( (p1.x)^2 + (p1.y)^2+ (p1.z)^2)
End Function

Function unit(p1 As pt) As pt  '=normalize
    Dim As Single L=length(p1)
    Return Type(p1.x/L,p1.y/L,p1.z/L)
End Function

#define dround(n,places) mid((str(int((n)*10^(places)+.5)/10^(places))),1,instr(ltrim(str((n)),"-"),".")+(places)+1)
#define onscreen (mx>0) and (mx<xres) and (my>0) and (my<yres)
#define incircle(cx,cy,radius,x,y) (cx-x)*(cx-x) +(cy-y)*(cy-y)<= radius*radius

Const pi=4*Atn(1)

Function cross(v1 As pt,v2 As pt) As pt '|cross product|= |v1| * |v2| *sin(angle between v1 and v2)
    Return Type(v1.y*v2.z-v2.y*v1.z,-(v1.x*v2.z-v2.x*v1.z),v1.x*v2.y-v2.x*v1.y)
End Function

Function getangle(p() As pt) As Single
    Dim As pt L1=(p(1).x-p(2).x,p(1).y-p(2).y)'leg 1
    Dim As pt L2=(p(3).x-p(2).x,p(3).y-p(2).y)'leg 2
    Dim As Single angle=Asin(length(cross(unit(L1),unit(L2))))*(180/pi)'angle between legs in degrees
    If cross(l1,l2).z>0 Then angle=360-angle
    Return angle
End Function


#macro display
angle=getangle(p())

Screenlock
Cls
Color 15
Draw String (20,20), "Drag circles by leftmouse button"
Color 3
Draw String (20,40), "Angle from red to green line (clockwise) in degrees"
Color 15
For z As Long=1 To 3
    Circle (p(z).x,p(z).y),5,7,,,,f
Next
Line(p(2).x,p(2).y)-(p(1).x,p(1).y),2
Line(p(2).x,p(2).y)-(p(3).x,p(3).y),4
Color 3
Draw String(p(2).x+20,p(2).y),Iif(Instr(dround(angle,2),"."),Rtrim(dround(angle,2),"0"),dround(angle,2))+chr(248)
Screenunlock
#endmacro

#macro mouse(m)
Dim As Long x=mx,y=my,dx,dy
While mb = 1
    Display():Sleep 1,1
    Getmouse mx,my,,mb
    If onscreen Then
        If mx<>x Or my<>y  Then
            dx = mx - x
            dy = my - y
            x = mx
            y = my
            p(m).x=x+dx
            p(m).y=y+dy
        End If
    End If
Wend
#endmacro

Screen 19
Dim As Integer xres,yres
Screeninfo xres,yres
Dim As pt p(1 To 3)={(100,100),(300,200),(200,400)} 'arbitary starting points
Dim As Integer mx,my,mb
dim as single angle
Do
   
    Getmouse(mx,my,,mb)
    display:Sleep 1,1
    For n As Long=1 To 3
        If incircle(p(n).x,p(n).y,10,mx,my) And mb=1 Then
            mouse(n)
        End If
    Next n
Loop Until Len(Inkey)





[dodicat]

I expect both to move in y-direction only while rotating.

You may be right with an object hitting a Borg ship. The boat and it's motor interacts with the water it is not skidding along a frictionless frozen lake with a rocket for propulsion.
https://www.youtube.com/watch?v=qaIqanZ00uo
However if there was a rocket engine on the block it would turn with the block and thus change the direction of the force in absolute space?
So to test this you would need to have a rocket of some kind (water bottle with vinegar and baking soda dissolved in water?) and a low friction setup such as a disc on an ice rink.

A sailing ship can move into the wind (Move towards the force)
If you put a spoon on a level table (curved side up) and force directly down on it with a pencil, the spoon will move at 90 degrees along the table to your force.
If you now put a book under the table thus tilting it up, and force the pencil directly down on the spoon will now ride up the table.
So the spoon has a motion component in the opposite direction to the force moving it.

But these two examples need a third interaction, the sea and the table.
I don't think you have a third thing in space.
On your own out there you could be stuck in the one place no matter how much you thrashed around.

[badidea]

However if there was a rocket engine on the block it would turn with the block and thus change the direction of the force in absolute space?
So to test this you would need to have a rocket of some kind (water bottle with vinegar and baking soda dissolved in water?) and a low friction setup such as a disc on an ice rink.

Yes, I understand that the direction eventually changes, but the key part of my original question was an thruster firing shortly. The reason for this is that the game will run in small time steps (milliseconds). For each small time step the physics need to be correct. The game does not know yet what the situation will be in the next time step.

With Joshy's 'd' I think I can solve the problem, roughly:
* Add center mass + the apparent mass (by the moment of inertia and this distance d)
* Determine a and (acceleration and velocity vector)where the force is acting
* Determine change in position and rotation
* Next time step
Something like that, to be worked out further...

I don't think you have a third thing in space.

Certainly, no tables and spoons in space :-)

[BasicCoder2]

On your own out there you could be stuck in the one place no matter how much you thrashed around.

https://www.youtube.com/watch?v=960CsCg ... be&t=1m44s
https://www.youtube.com/watch?v=RtWbpyjJqrU

[badidea]

Video Game Physics Tutorial - Part I: An Introduction to Rigid Body Dynamics

Looks like a useful resource. I will be camping for the next 2 weeks, so don't expect much coding progress from me, but I can at least read this on my phone and make some notes.

[dodicat]

Happy camping.

Code: Select all

 

Const pi=4*Atn(1)

Const BoxSize=100
Const systemconstant=1/boxsize

#define incircle(cx,cy,radius,x,y) (cx-x)*(cx-x) +(cy-y)*(cy-y)<= radius*radius
#define onscreen (mx>0) and (mx<xres) and (my>0) and (my<yres)

Type pt
    As Double x,y,z
End Type
Redim Shared As pt msg()

Operator -(p1 As pt,p2 As pt) As pt
Return Type(p1.x-p2.x,p1.y-p2.y)
End Operator

Operator +(p1 As pt,p2 As pt) As pt
Return Type(p1.x+p2.x,p1.y+p2.y)
End Operator

Operator *(f As Single,p As pt) As pt
Return Type(f*p.x,f*p.y)
End Operator

Function length(p1 As pt,p2 As pt) As Single
    Dim As Single diffx=p1.x-p2.x,diffy=p1.y-p2.y
    Return Sqr(diffx*diffx+diffy*diffy)
End Function

Function unit(p1 As pt) As pt
    #define lngth(p)  Sqr( (p1.x)^2 + (p1.y)^2+ (p1.z)^2)
    Dim As Single L=lngth(p1)
    Return Type(p1.x/L,p1.y/L,p1.z/L)
End Function

Function dot(v1 As pt,v2 As pt) As Single 'dot product |v1| * |v2| *cos(angle between v1 and v2)
    Return v1.x*v2.x+v1.y*v2.y+v1.z*v2.z
End Function

Function cross(v1 As pt,v2 As pt) As pt '|cross product|= |v1| * |v2| *sin(angle between v1 and v2)
    Return Type(v1.y*v2.z-v2.y*v1.z,-(v1.x*v2.z-v2.x*v1.z),v1.x*v2.y-v2.x*v1.y)
End Function

Function getangle(p() As pt) As Single
    Dim As pt L1=(p(1).x-p(2).x,p(1).y-p(2).y)'leg 1
    Dim As pt L2=(p(3).x-p(2).x,p(3).y-p(2).y)'leg 2
    Dim As Single angle=Acos(dot(unit(L1),unit(L2)))*(180/pi)'angle between legs in degrees
    If cross(l1,l2).z>0 Then angle=360-angle
    Return angle
End Function

'=========================
Function inpolygon(p1() As Pt,Byval p2 As Pt) As Long
    #macro Winder(L1,L2,p)
    ((L1.x-L2.x)*(p.y-L2.y)-(p.x-L2.x)*(L1.y-L2.y))
    #endmacro
    Dim As Integer index,nextindex,k=Ubound(p1)+1,wn
    For n As Integer=1 To Ubound(p1)
        index=n Mod k:nextindex=(n+1) Mod k
        If nextindex=0 Then nextindex=1
        If p1(index).y<=p2.y Then
            If p1(nextindex).y>p2.y Andalso  Winder(p1(index),p1(nextindex),p2)>0 Then wn+=1
        Else
            If p1(nextindex).y<=p2.y Andalso Winder(p1(index),p1(nextindex),p2)<0 Then wn-=1
        End If
    Next n
    Return wn
End Function

Sub drawpolygon(b() As pt,p() As Pt,rotmsg() As pt,Byref col As Ulong,Byval im As Any Pointer=0)
    Dim k As Long=Ubound(p)+1
    Dim As Long index,nextindex
    Dim As Long cx,cy
    For n As Long=1 To Ubound(p)
        cx+=p(n).x:cy+=p(n).y
        index=n Mod k:nextindex=(n+1) Mod k
        If nextindex=0 Then nextindex=1
        Line im,(p(index).x,p(index).y)-(p(nextindex).x,p(nextindex).y),col
    Next
    cx/=Ubound(p):cy/=Ubound(p)
    Paint (cx,cy),col,col
    For n As Long=Lbound(rotmsg) To Ubound(rotmsg)
        Pset(rotmsg(n).x,rotmsg(n).y),rotmsg(n).z'rgb(200,0,0)
    Next n
End Sub

Function rotate(pivot As pt,p As pt,a As Single,d As Single=1) As pt
    Return Type<pt>(d*(Cos(a)*(p.x-pivot.x)-Sin(a)*(p.y-pivot.y)) +pivot.x,_
    d*(Sin(a)*(p.x-pivot.x)+Cos(a)*(p.y-pivot.y)) +pivot.y)
End Function

Sub setUpPolygon(b() As pt,sz As Long,x As Long,y As Long,num As Long)
    Redim b(1 To num)
    Dim As Long ctr
    For n As Single=0 To 3*pi Step 2*pi/num
        ctr+=1
        If ctr>num Then Exit For
        b(ctr).x=x+sz*Cos(n)
        b(ctr).y=y+sz*Sin(n)
    Next n
End Sub

Sub turnpolygon(b() As pt,rot() As pt,rotmsg() As pt,a As Single,f As pt,translate As pt)
    For n As Long=Lbound(rot) To Ubound(rot)
        rot(n)= rotate(f,b(n),a)
        rot(n)+=translate
    Next n
    For n As Long=Lbound(rotmsg) To Ubound(rotmsg)
        rotmsg(n)= rotate(f,msg(n),a)
        rotmsg(n)+=translate
        rotmsg(n).z=msg(n).z
    Next n
   
End Sub

Function movepolygon(b() As pt,msg() As pt,dirn As pt) As pt
    Dim As pt p
    For n As Long=Lbound(b) To Ubound(b)
        b(n).x+=dirn.x
        b(n).y+=dirn.y
        p.x+=b(n).x
        p.y+=b(n).y 'get centre
    Next
    For n As Long=Lbound(msg) To Ubound(msg)'move message
        msg(n).x+=dirn.x
        msg(n).y+=dirn.y
    Next
   
    Dim As Long size=Ubound(b)-Lbound(b)+1
    Return Type(p.x/size,p.y/size)
End Function

Function shortline(fp As pt,p As pt,Ln As Long) As pt 'line of ln length
    Dim As Single diffx=p.x-fp.x,diffy=p.y-fp.y
    Dim As Single L=Sqr(diffx*diffx+diffy*diffy)
    Return Type(fp.x+Ln*diffx/L,fp.y+Ln*diffy/L)
End Function

Function drawline(x As Long,y As Long,angle As Single,dist As Long) As pt'line by angle
    Var x2=x+dist*Cos(angle)
    Var y2=y+dist*Sin(angle)
    Return Type(x2,y2)
End Function
'all the lines
Function lineto(c As pt,_out As pt,p2() As pt,dirn As pt,Byref strike As Long,b() As pt) As Single
    Var L=length(p2(1),p2(2))
    Var ang=Atan2(p2(2).y-p2(1).y,p2(2).x-p2(1).x)
    Dim As pt p=p2(1)
    Dim As Long ic,ctr
    Do
        p=shortline(p,p2(2),1)
        ic=inpolygon(b(),p)
        ctr+=1
        If ic Then strike=1 Else strike=0
    Loop Until ic Or ctr>L 'when it hits the box at p
   
    Line(p.x,p.y)-(p2(1).x,p2(1).y),Rgb(0,200,0) 'surface to the red point
   
    If strike Then  Line(c.x,c.y)-(p.x,p.y)
   
    Dim As pt eq=drawline(c.x,c.y,ang,-200)
    Dim As pt q=Type(c.x,c.y)
    ctr=0
    Do
        q=shortline(q,eq,1)
        ic=inpolygon(b(),q)
        ctr+=1
    Loop Until ic=0 Or ctr>400 'when exits the box at q
    If strike Then Line(c.x,c.y)-(q.x,q.y)
    Dim As pt pp(1 To 3)={c+(c-q),c,c+(c-p)} 'array to hold three points (two legs metting at t)
    _out=p          'out the intersection wth box
    dirn=unit(p-c)  'out the direction
   
    Dim As pt d=.3333*(p+q+c)'to show the angle
    Dim As Single angle=getangle(pp())
    If strike Then Draw String(d.x,d.y),Str(Int(angle))
    Return angle*pi/180 'return in radians
End Function

Function Regulate(Byval MyFps As Long,Byref fps As Long=0) As Long
    Static As Double timervalue,_lastsleeptime,t3,frames
    Var t=Timer
    frames+=1
    If (t-t3)>=1 Then t3=t:fps=frames:frames=0
    Var sleeptime=_lastsleeptime+((1/myfps)-T+timervalue)*1000
    If sleeptime<1 Then sleeptime=1
    _lastsleeptime=sleeptime
    timervalue=T
    Return sleeptime
End Function

Sub HandleRain(rain As Any Ptr)
    Var max=100
    Static As Single xx(max),yy(max)
    For i As Long = 0 To max
        xx(i) = Rnd*1024
        yy(i) = Rnd*768
        Put(xx(i),yy(i)),Rain,Alpha,200+Rnd*50
    Next
End Sub

#macro display
flag2=0
Screenlock
Cls
Draw String(20,20),"Move the small circles by mouse"
Draw String(20,50),"Press spacebar for impulse"
Draw String(20,80),"Right mouse click to reset"
Draw String(20,110),"Angular speed " &abs(angularimpulse*180/pi)
Draw String(20,140),"Linear speed " &speed
Draw String(20,170),"framerate = " &fps,Rgb(0,200,0)

pivot=movepolygon(b(),msg(),Type(dx,dy))
turnpolygon(b(),rot(),rotmsg(),-turn,pivot,Type(dx,dy))
drawpolygon(b(),rot(),rotmsg(),Rgb(0,100,255))'draw the polygon and return the centre
c=Type<pt>(xres\2,yres\2) 'centre

If flag=0  Then
    Locate 15
    Print "     weight "; impulse
    flag2=1
    a=  lineto(c,_out,p2(),dirn,strike,rot()) 'a=angle return as shown in radians
End If

If Len(Inkey) And flag2 And strike Then
    flag=1
    linearimpulse=impulse*Cos(a)/mass
    If a>pi/2 And a<3*pi/2 Then dirn=-1*dirn
    dx=-dirn.x*linearimpulse
    dy=-dirn.y*linearimpulse
    angularimpulse=impulse*Sin(a)/anginertia
    speed=Sqr(dx*dx+dy*dy) 'to show value on screen
End If

For z As Long=1 To 2 'the two drag spots
    Dim As Ulong clr=Iif(z=1,Rgb(200,0,0),Rgb(0,200,0))
    Circle (p2(z).x,p2(z).y),5,clr,,,,f
Next

If flag=0 Then impulse=length(p2(1),_out) 'arbitrary weight as a function of mouse potition

turn+=angularimpulse
handlerain(rain)

Screenunlock
Sleep regulate(60,fps),1
'====================
#endmacro

#macro mouse(m)
Scope
    Dim As Long x=mx,y=my,dx2,dy2
    While btn = 1 Or btn=2
        Display
        Getmouse mx,my,,btn
        If onscreen Then
            If mx<>x Or my<>y  Then
                dx2 = mx - x
                dy2 = my - y
                x = mx
                y = my
                p2(m).x=x+dx2
                p2(m).y=y+dy2
            End If
        End If
    Wend
End Scope
#endmacro

Sub MakeMsg(msg() As pt,xres As Long,yres As Long)
    Cls
    Redim msg(0)
    Print " Gone"+Chr(10)+"Fishing"
    Dim As Long ctr
    Var c=Point(1,1)
    For x As Long=0 To 60
        For y As Long=0 To 45
            If Point(x,y)<>c Then
                ctr+=1
                Redim Preserve msg(1 To Ubound(msg)+1)
                msg(Ubound(msg))=Type(x+xres/2,y+yres/2,Rgb(.75*ctr,255-.75*ctr,ctr/6))
               
            End If
        Next
    Next
End Sub

Function pickanumber As Long
    Cls
    Locate 12
    Print "click a number"
    #define map(a,b,_x_,c,d) ((d)-(c))*((_x_)-(a))/((b)-(a))+(c)
    Dim As Long mx,my,mb,ret
    Do
        Getmouse mx,my,,mb
        Locate 15
        If my>220 And my<235 Then
            Color ,Rgb(100,0,0)
            var n=Int(map(5,655,mx,3,30))
            If mx<675 Then Print Tab(90);"Choose  ";n;" ?"
            If mb Then ret=map(0,660,mx,3,30)
        Else
            Color ,Rgb(0,0,0)
        End If
        Locate 15
        For n As Long=3 To 30
            Print Right("0"+Str(n),2)+" ";
        Next
        Print
        sleep 10
    Loop Until ret
    Color ,Rgb(0,0,0)
    Return ret
End Function


Screen 20,32
Dim As Integer xres,yres
Screeninfo xres,yres
Dim As Single dx,dy,a,turn,angularimpulse,impulse,linearimpulse,speed

Dim As Single mass
Dim As Single anginertia
Dim As Long mx,my,btn,flag,flag2,fps,strike

Dim As pt c

Dim As pt _out,p2(1 To 2)={(200,200),(650,500)},dirn

Redim As pt b()

Dim As Long numsides

start:
numsides=pickanumber
                       'area of a regular polygon
 mass= systemconstant*boxsize^2*numsides*sin(2*pi/numsides)/2
                        'moment of inertia of a regular polygon
anginertia=systemconstant*((mass*boxsize^2)/6)*(1+2*(Cos(pi/numsides)^2))

MakeMsg(msg(),xres-40,yres-40)'create a string message
setUpPolygon(b(),boxsize*Sqr(2),xres\2,yres\2,numsides)

Redim As pt rot(Lbound(b) To Ubound(b))       'to hold rotated shape
Redim As pt rotmsg(Lbound(msg) To Ubound(msg))''to hold rotated message
Dim As pt pivot
Dim As Any Ptr Rain=Imagecreate(5,20)
Line rain,(0,0)-(5,20),Rgb(50,50,55)
Do
    Getmouse mx,my,,btn
    strike=0
    display
   
    For n As Long=1 To 2
        If incircle(p2(n).x,p2(n).y,10,mx,my) And btn=1 And flag2=1  Then
            mouse(n)
        End If
    Next n
   
    If btn=2 Or pivot.x<BoxSize  Or pivot.x>xres-BoxSize Or pivot.y<BoxSize Or pivot.y>yres-BoxSize Then 'reset
        Screenunlock
        sleep 1
        dx=0
        dy=0
        a=0:flag=0
        angularimpulse=0
        speed=0
        turn=0
        Goto start
    End If
Loop Until Multikey(1)
Imagedestroy rain
Sleep



[deltarho[1859]]

@dodicat

I am having problems running some of your code. Your very first code is working fine as is you code where we can drag & drop one of three circles. However, every other code where we should be able to drag & drop either the red or green circle sees the CPU max out on a core, I have four plus hyperthreading, and the Windows busy icon is displayed. The Task Manager displays 'not responding'. Eventually the system settles down but the circle that I dragged does not end up where I expected it to be and sometimes is off-screen.

I would have a bash at debugging but my graphics knowledge is woefully poor.

If anyone else is having issues please shout up so that we can eliminate my machine from any enquiries.

Ta much. <smile>