0
$\begingroup$

Is it possible to define a polar equation for a k-leaf rose with an inner radius for a k-leaf rose (as in this image)? I'm familiar with the general equation for a k-leaf rose $$r = \cos(k*\theta)$$ and the corresponding Cartesian equations $$x = \cos(k*\theta) * \cos(\theta)$$ and $$y = \cos(k*\theta) * \sin(\theta)$$ However, I've been unable to use these to come up with an equation that produces a rose with a hollow center. I'm very curious to know if it's possible. Thanks in advance, any insights will be greatly appreciated!

$\endgroup$
  • $\begingroup$ try $r=\cos(k\theta)+c$ where $c>1$ $\endgroup$ – David Quinn Jan 5 '16 at 11:32
0
$\begingroup$

You have two issues: cosine goes to zero, so the inner radius is not there, and cosine goes negative so adding a bit doesn't do what you want. If you square the cosine you can add a bit. This doesn't look like your figure, but it does have an inner radius

enter image description here If you increase the denominator it looks like this

enter image description here You can play with the parameters a bit

$\endgroup$
  • $\begingroup$ Awesome, thanks for the answer! It looks like increasing the additive constant to a number greater than 1 yields a smooth inner radius. $\endgroup$ – Ian Speers Jan 5 '16 at 18:13
  • $\begingroup$ @IanSpeers: I had fun with it. Just adding a larger constant makes the ratio of outer/inner too small to my eye. If you make the additive constant greater than $1$ you can remove the square from the cosine See how you like it. This goes fully to David Quinn's suggestion $\endgroup$ – Ross Millikan Jan 5 '16 at 18:55
  • $\begingroup$ Beautiful! The removal of the square will help to cut down on compute time. Now I just need to convert this bad-boy into Cartesian. Hopefully my calc isn't too rusty :). Thanks again! $\endgroup$ – Ian Speers Jan 5 '16 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.