1
$\begingroup$

I trying to make something like this (callout title. like in image) with animation nodes. And i need to draw line with thickness and animate it. If i use curve with Bevel Object (second curve) it give ugly result near twist (image1). I know how to draw curve or line mesh using animation nodes but they havent thickness. Does anyone know how to add thickness?

enter image description here enter image description here

$\endgroup$
4
$\begingroup$

Let the thickness be $T$, the length of the base be $L$, the length of the other side be $W$, and the angle between the base and the other side is $\theta$. Then you should create a mesh composed of 6 vertices with the following vertices locations:

Figure

$(0,0)$, $(0,T)$, and $(L,0)$ are trivial. For the point $(L-T\cot{(\theta/2)}, T)$, the y location is trivial, the x location is computed by noting that $T\cot{(\theta/2)} = T\frac{\text{base}}{T} = \text{base}$, where $\text{base}$ is the base of the triangle composed from $(L,0)$, the point, and its projection on the x axis. For the point $(L+W\cos{(\theta)}, W\sin{(\theta)})$, $W\cos{(\pi - \theta)} = W\frac{\text{base}}{W} = \text{base}$ where $\text{base}$ is the base of the triangle composed of $(L,0)$, the point, and the projection of the point on the x axis. And $W\sin{(\pi - \theta)} = W\frac{\text{height}}{W} = \text{height}$ where $\text{height}$ is the height of the triangle composed of $(L,0)$, the point, and the projection of the point on the x axis. Finally, for $(L+\cos{(\pi-\theta)}(W-T), \sin{(\pi-\theta)}(W+T)))$, we add a vector to the aforementioned point, this vector is computed from the triangle composed from $(L-T\cot{(\theta/2)}, T)$, its projection on the the side whose length is $W$, and the intersection point between the projection lines of the point on the x axis and the side length whose length is $W$. Which conclude the derivation of the point locations.

$\endgroup$

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.