Tangent line to a 3D curve at a given 3D point

Question

I'm trying to compute tangent vector(x,y,z) to a 3D curve(start x1,y1,z1 end x2,y2,z2) at a point (x0,y0,z0). Can anyone tell if this information is sufficient to calculate the tangent vector? If yes can anyone tell the formula with the help of some values. Help is greatly appreciated!!!

Answer 1

Can anyone tell if this information is sufficient to calculate the tangent vector?

No. We need to know the shape of the curve.

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Knowing the curve passes via $\vec{p}_s = (x_s , y_s , z_s)$ and $\vec{p}_e = (x_e , y_e , z_e)$ does not tell us anything about the shape of the curve. In the simplest case, the curve would be a straight line, and in that case its tangent is everywhere the same, $\vec{p}_e - \vec{p}_s$.

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In computer programs, cubic Bézier curves are ubiquitous. They are defined using four points. The curve passes through the first point $\vec{p}_1 = (x_1 , y_1 , z_1) = \vec{p}_s$ and the last point $\vec{p}_4 = (x_4 , y_4 , z_4) = \vec{p}_e$, but the other two just control the shape of the curve. Mathematically, $$\bbox{ \begin{cases} x(t) = (1-t)^3 x_1 + 3 (1-t)^2 t x_2 + 3 (1-t) t^2 x_3 + t^3 x_4 \\ y(t) = (1-t)^3 y_1 + 3 (1-t)^2 t y_2 + 3 (1-t) t^2 y_3 + t^3 y_4 \\ z(t) = (1-t)^3 z_1 + 3 (1-t)^2 t z_2 + 3 (1-t) t^2 z_3 + t^3 z_4 \\ \end{cases}, \quad 0 \le t \le 1 } \tag{1}\label{NA1}$$ Note that that notation is just a good, easy form to express the curve. We could write the curve coordinates as third-degree polynomials, $$\bbox{ \begin{cases} x(t) = X_0 + t X_1 + t^2 X_2 + t^3 X_3 \\ y(t) = Y_0 + t Y_1 + t^2 Y_2 + t^3 Y_3 \\ z(t) = Z_0 + t Z_1 + t^2 Z_2 + t^3 Z_3 \\ \end{cases} }, \quad \bbox{\begin{cases} X_0 = x_1 \\ X_1 = 3 x_2 - 3 x_1 \\ X_2 = 3 x_3 - 6 x_2 + 3 x_1 \\ X_3 = x_4 - 3 x_3 + 3 x_2 - x_1 \\ \end{cases} } \tag{2}\label{NA2}$$ but in computer programs $\eqref{NA1}$ is more numerically stable.

Non-uniform rational B-splines are a more general form of curves, used especially in modeling. You can obtain the tangent vectors on NURBS using basically the same way as below; you just use the NURBS equations for the coordinates instead.

If you've ever used Inkscape to doodle shapes, you probably have used the Bézier curve tool to draw cubic Bézier curves already. Start and end points are shown as boxy dots, and those control points as "handles", a circular dot connected to the start or end point on the curve. They are used in PostScript and SVG formats as well.

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The tangent vector at any point $t$ along the curve is simply the derivative of the coordinates. Using $\eqref{NA2}$ above, $$\bbox{ \begin{cases} \displaystyle \frac{d x(t)}{d t} = X_1 + 2 X_2 t + 3 t^2 X_3 \\ \displaystyle \frac{d y(t)}{d t} = Y_1 + 2 Y_2 t + 3 t^2 Y_3 \\ \displaystyle \frac{d z(t)}{d t} = Z_1 + 2 Z_2 t + 3 t^2 Z_3 \\ \end{cases} } \tag{3}\label{NA3}$$

However, we can also describe the tangent in quadratic Bézier form, by deriving $\eqref{NA1}$, and reordering the terms. We get $$\bbox{ \begin{cases} \displaystyle \frac{d x(t)}{d t} = (1-t)^2 ( 3 x_2 - 3 x_1) + 2 (1-t) t ( 3 x_3 - 3 x_2 ) + t^2 ( 3 x_4 - 3 x_3 ) \\ \displaystyle \frac{d y(t)}{d t} = (1-t)^2 ( 3 y_2 - 3 y_1) + 2 (1-t) t ( 3 y_3 - 3 y_2 ) + t^2 ( 3 y_4 - 3 y_3 ) \\ \displaystyle \frac{d z(t)}{d t} = (1-t)^2 ( 3 z_2 - 3 z_1) + 2 (1-t) t ( 3 z_3 - 3 z_2 ) + t^2 ( 3 z_4 - 3 z_3 ) \\ \end{cases} } \tag{4}\label{NA4}$$ This form, $\eqref{NA4}$, is again numerically quite stable, and is very suitable for use in a computer program.

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That leaves one more problem to solve: how to find the curve parameter $t$ that corresponds to a point I already have?

There are algebraic solutions for linear curves (lines), quadratic curves, and cubic curves. Essentially, you can pick one of the coordinate functions, say $x$, and solve $$x(t) - x = 0$$ using any root-finding algorithm. For quadratic curves, you find up to two real $t$; for cubic, up to three real $t$. You ignore non-real roots and $t \lt 0$ and $t \gt 0$, and check the candidates left if $y(t) = y$ and $z(t) = z$ at sufficient precision.

In computer programs, you always do such tests as $\lvert y(t) - y \rvert \le \epsilon$ and $\lvert z(t) - z \rvert \le \epsilon$, where $\epsilon$ represents the precision in your coordinates: the largest difference in coordinates that you consider neglible. For example, if you read your point cloud coordinates with three decimals, use $\epsilon = 0.0005$.

If your computer program already implements a fast curve point evaluation function (the simple form is fast enough!), a binary search in $t$ may prove faster. (It is also a very interesting approach if you have many curve types, or use polynomial functions of possibly very high degree for your curves.) The only "trick" needed is that because some curves have loops (i.e., the coordinates are not monotonic functions), you may have to split the initial $0 \le t \le 1$ to smaller ranges, to ensure you find the $t$ on the curve closest to the specified point, rather than a value of $t$ that is closer than its nearby points. (It is the difference between local and global minima.)

Essentially, you pick one of the coordinate functions, a range $0 \le t_{min} \lt t_{max} \le 1$ in which the function is monotonic (increases or decreases, but not both), and then

Function BSearch(coordfunc, target, tmin, tmax, epsilon):
    Let  vmin = coordfunc(tmin) - target
    Let  vmax = coordfunc(tmax) - target
    If vmin > vmax:
        Let  ttmp = tmin ; tmin = tmax ; tmax = ttmp
        Let  vtmp = vmin ; vmin = vmax ; vmax = vtmp
    End If
    If vmin > 0:
        Return tmin
    Else If vmax < 0:
        Return tmax
    End If

    Loop:
        Let  t = (tmin + tmax) / 2
        If vmax - vmin <= epsilon:
            Return  t
        End If

        Let  v = coordfunc(t) - target
        If (v < 0) and (v >= vmin):
            tmin = t
        Else
        If (v > 0) and (v <= vmax):
            tmax = t
        Else
        If (v == 0):
            Return t
        Else:
            # Function is non-monotonic.
            Return Failed
        End If
    End Loop
End Loop            

The function will return the value of t that is closest to the target coordinate, within the range specified. If it detects the function is non-monotonic, it will return Failed. Because the function is only sampled, it does not reliably detect non-monotonic functions, and can return a non-optimal t for such functions.