# Heart

Shape · (x, y, z) = f(u, v, t)

`(0.1 · sin(v) · (15 · sin(u) − 4 · sin(3u)),  0.1 · sin(v) · (19 · cos(u) − 5 · cos(2u) − 2 · cos(3u)),  0.1 · 8 · cos(v))`

[Open in the app](https://www.wavelace.com/app#p=57) · [This page](https://www.wavelace.com/presets/heart)

### What it draws

The parameter `u` goes round the outline of a heart, and `v` sweeps that outline from the back of the figure to the front. The second expression is the height, `0.1·sin(v)·(19·cos(u) − 5·cos(2u) − 2·cos(3u))`. The first is the width, `0.1·sin(v)·(15·sin(u) − 4·sin(3u))`, and the third, `0.8·cos(v)`, is the depth. Nothing depends on `t`, so the figure stands still.

At `v = π/2` the factor `sin(v)` is 1, and the outline is the heart at full size in the vertical plane `z = 0`. Its point is at `u = π`, height `−2.2`. The cleft between the lobes is at `u = 0`, height `1.2`, and the lobes crest at `1.49`. The widest points, `x = ±1.9`, fall at `u = π/2` and `3π/2`.

### The heart curve

Both outline expressions are *trigonometric polynomials*: sums of sines and cosines of whole multiples of `u`. In the width, `15·sin(u) − 4·sin(3u)` is a fattened form of `16·sin(u)³`, which expands exactly to `12·sin(u) − 4·sin(3u)`. The extra 3 widens the lobes from `±1.6` to `±1.9`.

In the height, `19·cos(u)` alone would draw an ellipse. The term `−5·cos(2u)` pulls the top down at `u = 0`, and `−2·cos(3u)` pushes the two lobes up on either side. Between them they are the whole dent.

### The sweep

The outline is scaled by `sin(v)` while it is placed in depth at `0.8·cos(v)`, and that pair traces an ellipse as `v` runs `0 → π`. The heart therefore grows from a point at the back, reaches full size in the middle plane, and shrinks to a point at the front. It is a solid built the way a sphere is, with heart-shaped sections in place of circles.

### Try

- `Front` in the deck: straight on, the outline is the plane heart curve.
- Put the textbook cube in the width, `0.1·sin(v)·16·sin(u)³`: the same heart with narrower lobes, `±1.6` wide.
- Set `Span of v (×π)` to 2. Past `v = π` the factor `sin(v)` is negative, so a second heart, point upwards, is traced through the first.
- Give the depth a beat, `0.1·8·cos(v)·(1 + 0.3·sin(t))`: the figure swells front to back every `2π ≈ 6.28` seconds at `Speed` 1.

### Read more

- [Heart Curve](https://mathworld.wolfram.com/HeartCurve.html)
- [Parametric surface](https://en.wikipedia.org/wiki/Parametric_surface)
- [Trigonometric polynomial](https://en.wikipedia.org/wiki/Trigonometric_polynomial)