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Pomax
2018-06-24 18:02:32 -07:00
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5
components/sections/curvefitting/content.en-GB.md
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@@ -246,7 +246,4 @@ So let's try it out! The following graphic lets you place points, and will start
</Graphic>
</div>
You'll note there are also two convenient buttons: the "toggle" button lets you toggle between equidistance `t` values, and distance ratio along the polygon, and the "reset" button just clears the graphic so you can draw a new set of points. Arguably more interesting is that once you have points to abstract a curve, you also get <em>direct control</em> over the time values, because if the time values are our degree of freedom, you should be able to freely manipulate them and see what the effect on your curve is.
You'll note there are also two convenient buttons: the "toggle" button lets you toggle between equidistance `t` values, and distance ratio along the polygon, and the "reset" button just clears the graphic so you can draw a new set of points. Arguably more interesting is that once you have points to abstract a curve, you also get <em>direct control</em> over the time values, because if the time values are our degree of freedom, you should be able to freely manipulate them and see what the effect on your curve is.

6
components/sections/extremities/content.en-GB.md
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@@ -66,8 +66,10 @@ Quartic—fourth degree—curves have a cubic function as derivative. Now, cubic
generic form. So:
\[
very\ hard:\ solve\ at^3 + bt^2 + ct + d = 0\\
easier:\ solve\ t^3 + pt + q = 0
\begin{aligned}
\textit{very hard: solve } & at^3 + bt^2 + ct + d = 0 \\
\textit{easier: solve } & t^3 + pt + q = 0
\end{aligned}
\]
This is easier because for the "easier formula" we can use [regular calculus](http://www.wolframalpha.com/input/?i=t^3+%2B+pt+%2B+q) to find the roots (as a cubic function, however, it can have up to three roots, but two of those can be complex. For the purpose of Bézier curve extremities, we can completely ignore those complex roots, since our *t* is a plain real number from 0 to 1).