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Thread ID: 134380	2013-06-26 04:47:00	Drawing a parabola	Tony (4941)	PC World Chat

Post ID	Timestamp	Content	User
1347026	2013-06-26 04:47:00	OK, here's one for all you mathematicians/geometers. I suspect the solution is ridiculously easy and I just can't see it - or it's not. I understand all about how a parabola is a graphical representation of a quadratic equation, but I don't have a particular equation in mind. What I effectively want to do is to draw a rectangle, and then have a parabolic curve going from a corner to the centre of the opposite side and over to the other opposite corner. Any suggestions? The tools I have available are: pencil and paper and compasses (least preferred option) Corel Draw Turbocad TIA	Tony (4941)
1347027	2013-06-26 05:17:00	The general equation of a parabola is y² = 4ax So use the rectangle to set the x and y coordinates. The middle of the side where the parabola touches tangentially set as x=0, y=0 . The corners where the parabola passes through will be x1, y1 say, which you get from the size of the rectangle. Hence you can find the value of 'a'. Hence you can calculate and plot as many points along the parabola as you wish. In the good old days I'd have joined up the points with a french curve. But I'm sure Turbocad will plot a smooth curve through the points for you. Edit:There may even be an option within Turbocad to plot a parabola of your choice if you specify the important points, like there would be to draw a circle.	Terry Porritt (14)
1347028	2013-06-26 05:33:00	There may even be an option within Turbocad to plot a parabola of your choice if you specify the important points, like there would be to draw a circle.That would be the easy way! I can't find a function that does that. All my version gives me are bezier curves and sections of a circle. There is something that allows you to plot a curve using control points which may do the trick. Thanks.	Tony (4941)
1347029	2013-06-26 05:50:00	Well a quadratic bezier is what you want, ie a parabola. You would just need to specify a few points as I mentioned above.	Terry Porritt (14)
1347030	2013-06-26 05:52:00	Google Images --> Ctrl + P? :D	Chilling_Silence (9)
1347031	2013-06-26 05:57:00	Google Images --> Ctrl + P?I'd have to find one with correct curve, wouldn't I?	Tony (4941)
1347032	2013-06-26 06:03:00	Well a quadratic bezier is what you want, ie a parabola. You would just need to specify a few points as I mentioned above.I didn't realize a quadratic bezier was a parabola:blush:. It's a long time since I did any of this stuff, and I'm not sure how to go about plotting the points. I see that I set up an x axis with positive going one way and negative the other, and the y axis is obviously the other side of the rectangle. I'm not sure what you are saying by "x1 y1" and how you derive the points from the formula.	Tony (4941)
1347033	2013-06-26 06:21:00	Let me see if I can describe what I mean without a drawing :) Draw a rectangle. The left hand vertical side is on the x=0 axis, and half way up that side we make y=0, ie y axis is vertical, x axis is horizontal. For the sake of argument suppose the rectangle has a width or height of 10 and a length of 20 The coordinates of the left hand corners will be x=0, y=5 at the top, and x=0 y=-5 at the bottom. The coordinates of the right hand side corners (which I called x1 y1 for just one corner) will be x=20 y=5, x=20 y=-5 The parabola will be tangential to the left hand side of the rectangle at x=0 y=0, and pass through the corners at the right hand side. So using y²=4ax and say the top right hand corner coordinates 5x5 = 4a x 20 a= 25/80 =0.3125 Hence points along the parabola can be calculated. Hopefully the bezier function in Turbocad will just require the these two points to be entered. I said previously the general equation of a parabola, I should have said the simplest form of the equation is y²=4ax If the origin of the parabola is not at x=0 y=0 then it will be slightly more complicated.	Terry Porritt (14)
1347034	2013-06-26 06:44:00	Thanks for that Terry. I won't be able to do any more till tomorrow, but I'll report back. BTW I found this: www.oberonplace.com and couldn't make any sense of that either. :)	Tony (4941)
1347035	2013-06-26 06:52:00	Thanks for that Terry. I won't be able to do any more till tomorrow, but I'll report back. BTW I found this: www.oberonplace.com and couldn't make any sense of that either. :) Their parabola is turned through 90 deg compared with the one I envisaged because they are using the equation y=ax² instead of y²=4ax	Terry Porritt (14)
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