Graphic Challenge

I’ve been looking at a Reuleaux Triangle (download below). Trying to create it in BASIC is certainly challenging – tried a complex parametric equation (off web) which fails when tested (bracket in the wrong place, maybe). Anyone fancy having ago at coding it?

http://www.riscosbasic.uk/problems/Reuleaux.zip

If this is the same shape as a 50 p piece but with three sides (i.e. has constant width when it rotates) then the following would do it, using MakeDraw:

x1=0
y1=200
x2=-100
y2=200-50*SQRT(3)
x3=100
y3=y2
PROCmdArcC(x1,y1,x3,y3,x2,y2)
PROCmdArcC(x3,y3,x2,y2,x1,y1)
PROCmdArcC(x2,y2,x1,y1,x3,y3)

Something like

DEF PROCtriangle(x1%, y1%, x2%, y2%)
LOCAL h%, v%, side, phi, x3%, y3%

h% = ABS(x1% - x2%)
v% = ABS(y1% - y2%)

phi = (PI / 3) + TAN(v% / h%)
side = SQR(h%^2 + v%^2)

x3% = x1% + (side * COS(phi))
y3% = y1% + (side * SIN(phi))

MOVE x1%, y1%
PLOT 21, x2%, y2%
PLOT 21, x3%, y3%
PLOT 21, x1%, y1%

PROCarc(x1%, y1%, x2%, y2%, x3%, y3%)
PROCarc(x2%, y2%, x3%, y3%, x1%, y1%)
PROCarc(x3%, y3%, x1%, y1%, x2%, y2%)
ENDPROC

DEF PROCarc(x%, y%, x1%, y1%, x2%, y2%)
MOVE x%, y%
MOVE x1%, y1%
PLOT 165, x2%, y2%
ENDPROC

where x1, y1% and x2%, y2% for PROCtriangle() are the coordinates of the ends of one side of the inner triangle?

That’s not a Reuleaux Triangle, it’s the rotor out of my car

Are they still making those, or do you have quite an old car? What model?

Are they still making those, or do you have quite an old car? What model?

I suspect Druck has/had a Mazda rotary, not the old NSU Ro80.

I’ve got an 2009 RX-8 R3, they stopped making them in 2012. I’m thinking of getting something more economical before the petrol ban comes in, may be a 4.4L V8.

Reply to Chris

I’ll certainly have a look at your suggestion. I’ve never used MakeDraw so would be a good time to investigate. I prefer a standalone program as I can ‘Chain’ others for a pseudo slideshow.

Interestingly Steve your program using coordinates to render shape near the centre of the screen draws almost the whole circle leaving out the convex part along the triangle edges (see below). Removing the triangle actually creates a rather nice pattern. The coordinates with smaller values renders an accurate Reuleaux triangle. Coordinates seem quite critical. What I need now is to be able to rotate the shape around the screen centre.
Thanks for your submission.

http://www.riscosbasic.uk/problems/Reuleaux_combo.zip

As it rotates, do you not need it also to shift side to side or up and down, like the Wankel engine’s rotor?

What I need now is to be able to rotate the shape around the screen centre.

A slightly different take:

DEFPROCReuleaux(radius,rotate)
 LOCAL r(),rotate()
 DIM r(2,1),rotate(1,1)
 rotate()=COSrotate,SINrotate,-SINrotate,COSrotate 
 sinrad=radius*SIN(PI/6):REM =radius/2
 cosrad=radius*COS(PI/6):REM =sinrad*SQR3
 r()=0,radius,-cosrad,-sinrad,+cosrad,-sinrad
 r()=r().rotate()
 MOVE r(0,0),r(0,1)
 DRAW r(1,0),r(1,1)
 DRAW r(2,0),r(2,1)
 DRAW r(0,0),r(0,1)
 REM MOVE r(0,0),r(0,1) not required with triangle
 MOVE r(1,0),r(1,1)
 PLOT 165,r(2,0),r(2,1)
 MOVE r(0,0),r(0,1)
 PLOT 165,r(1,0),r(1,1)
 MOVE r(2,0),r(2,1)
 PLOT 165,r(0,0),r(0,1)
 CIRCLE 0,0,radius
ENDPROC

This draws the triangle, the reuleaux and the circumscribing circle about the ORIGIN, which must be set. The rotation is anti-clockwise about the centre.

As it rotates, do you not need it also to shift side to side or up and down

That happens here, but ought to be compensated for.

Here is my version of a BASIC program to draw the Reuleaux triangle using nested FOR loops instead of Procedures.
The constant A is the radius of each circular arc, constant B the distance of each vertex from the centre and N is the number of line segments approximating each arc. The numbers 640 and 512 are to get the origin, also the centre of the triangle, near the centre of the screen. The pair (C!, C2) is the coordinate pair for the centre of an arc.

I don’t know why, but I can’t get the third line to come out correctly: On the end should be (3), so that B is A divided by the square root of 3.

A = 200
N = 100
B = A/SQR

MODE 31
MOVE 640 – B/2, 512 + A/2

FOR M = 0 TO 2
Phi = 2*M*PI/3: Theta = 5*PI/6 + Phi
C1 = B*COS(Phi): C2 = B*SIN(Phi)
FOR K = 1 TO N
DRAW 640 + C1 + A*COS(Theta + K*PI/(3*N)), 512 + C2 + A*SIN(Theta + K*PI/(3*N))
NEXT K
NEXT M

END

This program can be easily generalised to ‘regular curvilinear polygons’ with any number of sides.
I think a program with just one FOR loop is possible.

Second attempt at my post above.

Here is my version of a BASIC program to draw the Reuleaux triangle using nested FOR loops instead of Procedures.
The constant A is the radius of each circular arc, constant B the distance of each vertex from the centre and N is the number of line segments approximating each arc. The numbers 640 and 512 are to get the origin, also the centre of the triangle, near the centre of the screen. The pair (C!, C2) is the coordinate pair for the centre of an arc

A = 200
N = 100
B = A/SQR(3)

MODE 31
MOVE 640 - B/2, 512 + A/2

FOR M = 0 TO 2
Phi = 2*M*PI/3: Theta = 5*PI/6 + Phi
C1 = B*COS(Phi): C2 = B*SIN(Phi)
FOR K = 1 TO N
DRAW 640 + C1 + A*COS(Theta + K*PI/(3*N)), 512 + C2 + A*SIN(Theta + K*PI/(3*N))
NEXT K
NEXT M

END

This program can be easily generalised to ‘regular curvilinear polygons’ with any number of sides.
I think a program with just one FOR loop is possible.

Many thanks for the coding – what an great response! One of the most interesting is Steve Drain’s example. Rotating the shape is simple. Only 2 parameters (radius and rotate) required. Wonder if animation would benefit from contra-rotating ‘Reuleaux Triangles’- I’ll try it.

Rotating the shape is simple.

BASIC’s array operations and ORIGIN are your friends. ;-)

The example I posted was specific for 3 sides. There is a generalised set of routines at:

Have fun.

Here is another program that includes animation.

MODE 31
w = 120 : REM w is the half-width of the guide.
b = 2*w/SQR(3)
theta = 0
S = -41.89
H = 511 - b

MOVE 640-w,0
DRAW 640-w,1023
MOVE 640+w,0
DRAW 640+w,1023

DIM V0(1,2), V(1,2), R(1,1), F(1,2), L(1,2)
V0() = 0, -w, w, b, -b/2, -b/2
L() = 640, 640, 640, 512, 512, 512

V() = 640, 640-w, 640+w, 1023, 1023-3*b/2, 1023-3*b/2

MOVE V(0,0),V(1,0)
DRAW V(0,1),V(1,1)
DRAW V(0,2),V(1,2)
DRAW V(0,0),V(1,0)

MOVE V(0,1),V(1,1)
PLOT 165,V(0,2),V(1,2)
MOVE V(0,0),V(1,0)
PLOT 165,V(0,1),V(1,1)
MOVE V(0,2),V(1,2)
PLOT 165,V(0,0),V(1,0)

PROCwait(3)

FOR I=0 TO 18

MOVE 640-w,0
DRAW 640-w,1023
MOVE 640+w,0
DRAW 640+w,1023

Y = S * I + H
theta = I*PI/9

M = FNquot(theta,PI/3) : phi = FNremain(theta,PI/3)
REM alpha = 2*M*PI/3: beta = 5*PI/6 + alpha

E = (-1)^M * (w - b* COS(phi-PI/6))

R() = COS(theta), -SIN(theta), SIN(theta), COS(theta)

V() = R().V0()
V() = V() + L()

REM This block plots the diagram without the correction for the disc centre.
REM MOVE V(0,0),V(1,0)
REM DRAW V(0,1),V(1,1)
REM DRAW V(0,2),V(1,2)
REM DRAW V(0,0),V(1,0)

F() = E,E,E,Y,Y,Y
V() = V() + F()

MOVE V(0,0),V(1,0)
DRAW V(0,1),V(1,1)
DRAW V(0,2),V(1,2)
DRAW V(0,0),V(1,0)

MOVE V(0,1),V(1,1)
PLOT 165,V(0,2),V(1,2)
MOVE V(0,0),V(1,0)
PLOT 165,V(0,1),V(1,1)
MOVE V(0,2),V(1,2)
PLOT 165,V(0,0),V(1,0)

PROCwait(1)
CLG

NEXT I

END

DEF FNquot(T,D) = INT(T/D)
DEF FNremain(T,D) = T - INT(T/D) * D
DEF PROCwait(delay)
TIME = 0
REPEAT
UNTIL TIME >= 100*delay
ENDPROC

The program needs tidying up, but it does show the Reuleaux triangle rolling
down a slot. The vertical slot is represented by the two vertical lines, so
it’s the side view. The rather precise looking value for the variable S is
the value such that there is no skidding along the sides of the slot. With
adjustment of a few values the animation could be made smoother.