[A83] Re: The shape of a cord
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[A83] Re: The shape of a cord
You can't exactly do that...first of all l = 5, and (x1, y1) and (x2,
y2) have to be at points exactly 5 units away on this parabola...
-----Original Message-----
From: assembly-83-bounce@lists.ticalc.org
[mailto:assembly-83-bounce@lists.ticalc.org] On Behalf Of Sebastiaan
Roodenburg
Sent: Wednesday, November 28, 2001 5:32 PM
To: assembly-83@lists.ticalc.org
Subject: [A83] Re: The shape of a cord
>
>
> Would you mind telling us what x1, y1, x2, and y2 are?
>
Does that matter?? Take some random values, try:
x1 = 2
x2 = 8
y1 = 2
y2 = 4
l = 7
x3 = 5
>
> -----Original Message-----
> From: assembly-83-bounce@lists.ticalc.org
> [mailto:assembly-83-bounce@lists.ticalc.org] On Behalf Of Patai
Gergely
> Sent: Wednesday, November 28, 2001 4:53 PM
> To: assembly-83@lists.ticalc.org
> Subject: [A83] Re: The shape of a cord
>
>
> > Not quite sure I understand, do you want to know how much
> cord it will
> > take to make a parabola-shaped cord go 5 feet from (x1,
> y1) to (x2, y2)
> > and still touch (x3, y3)?
>
> Let's go through it again. We are in two dimensions.
> I have a cord of known length and given the coordinates of
> its endpoints, I'd like to find approximate its shape
> with a parabola. To set up the equation for this
> parabola, I need a third point, whose x coordinate
> is also fixed to the average of the other two x
> values. Some ASCII art:
>
> Q(x2,y2)
> P(x1,y1) /
> \ /
> \ / <--- the length of this parabola segment
> \__/ is l
> R(x3,y3)
>
> We know x1, y1, x2, y2, l, and we fix x3 to be
> (x1+x2)/2. Question: how much is y3? Normally there
> should be two solutions, if l is greater than
> the distance of P and Q, but let that not disturb
> us for now, we are clever enough to choose the
> right one if we have both. :)
>
> PG
>
>
>
>
>
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