# Thread:

1. ## Calculating Normal Acceleration

Hi guys,

I have a 3D curve that is defined by a parameterization in the form of x(t), y(t), and z(t). In the context of the problem the curve is a waterslide and I need to calculate the normal acceleration at various points along the curve. I can calculate the velocity at any point by a simple conservation of energy from the top of the slide to the point of interest, but what do I need to do to calculate normal acceleration?

2. You have x(t), y(t), z(t).

Velocity vector v = (dx/dt,dy/dt,dz/dt).
Acceleration vector a=dv/dt = $(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2},\frac{d^2z}{d t^2})$.
v direction tangential to the curve.
Find vector $\tau$ = v/|v| |n|=1 tangential to the curve.
Tangential acceleration
$a_t=a\tau$
Normal acceleration
${a_n}^2=a^2-{a_t}^2$.

3. Thank you for helping, but I must admit that I do not totally get it yet. Perhaps someone could help me with a simple example and then I will be able to apply it to my more complicated problem.

Imagine that my curve is defined by the simple parametrization:

x(t)=sin(t)
y(t)=cos(t)
z(t)=.01t

I am interested in the point where t=90 degrees, which represents one specific point on the curve. Based on the conservation of energy and the height at that point, I know that the rider will be traveling at 10m/s. How do I calculate his normal acceleration?

4. $r(t) = ( \sin t, \cos t, 0.01t )$

$r'(t) = \frac{dr(t)}{dt} = ( \cos t, -\sin t, 0.01 )$

$a(t) = r''(t) = \frac{d^2r(t)}{dt^2} = ( -\sin t, -\cos t, 0 )$

$T = \frac{r'(t)}{|r'(t)|}$ - unit vector tangential to the curve

$N = \frac{T'}{|T'|}$ - unit vector normal to the curve

$a(t) = Ta_T + Na_N$ where

$a_T = T\cdot a(t) = \frac{|r'(t) \cdot a(t)|}{|r'(t)|}$

$|r'(t)| = \sqrt{1.01}$.

$r'(t) \cdot a(t) = ( \cos t, -\sin t, 0.01 )( -\sin t, -\cos t, 0 ) = 0$ there is no tangetial component only normal, thus

$a(t) = Na_N = ( -\sin t, -\cos t, 0 )\: and\: |a(t)| = 1$.
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