6.3 Snap, Crackle, and Pop

First a word about terminology. We will need to introduce a few extra derivatives of the position. It would be fun to give them names with a reasonable `feel' to them, just like with jerk. What type of motion feels even more restless than jerking motion? A sudden snap comes to mind. A what changes its state more sudden than a snap -- how about a crackle? And for those familiar with American rice crispies culture, a pop cannot be far away, and indeed, if something pops it really changes high derivatives of positions in a substantial way! We are not making these names up: we have seen them used a few times before, although the precise source is likely to be lost in (recent) history. So here they are:

$${\bf s}= \frac{d^4}{dt^4} {\bf r}\qquad ; \qquad {\bf c}= \f... ...dt^5} {\bf r}\qquad ; \qquad {\bf p}= \frac{d^6}{dt^6} {\bf r}$$ (6.12)

snap, crackle, and pop, respectively.

We are now in a position to write the Taylor series for the four quantities that appear in Eqs. 6.1, 6.2, up to crackle:

$${\bf r}_{i+1} = {\bf r}_i + {\bf v}_i dt + {\textstyle\frac{1}{2}}{\bf a}_{i}(dt)... ...tyle\frac{1}{24}}{\bf s}_{i}(dt)^4 + {\textstyle\frac{1}{120}}{\bf c}_{i}(dt)^5$$ (6.13)

$${\bf v}_{i+1} = {\bf v}_i + {\bf a}_i dt + {\textstyle\frac{1}{2}}{\bf j}_{i}(dt)... ...tstyle\frac{1}{6}}{\bf s}_{i}(dt)^3 + {\textstyle\frac{1}{24}}{\bf c}_{i}(dt)^4$$ (6.14)

$${\bf a}_{i+1} = {\bf a}_i + {\bf j}_i dt + {\textstyle\frac{1}{2}}{\bf s}_{i}(dt)^2 + {\textstyle\frac{1}{6}}{\bf c}_{i}(dt)^3$$ (6.15)

$${\bf j}_{i+1} = {\bf j}_i + {\bf s}_i dt + {\textstyle\frac{1}{2}}{\bf c}_{i}(dt)^2$$ (6.16)

We can now eliminate snap and crackle at time $t_i$, expressing them in terms of the acceleration and jerk at times $t_i$ and $t_{i+1}$, using Eqs. 6.15, 6.16. We find:

$${\bf s}_i = -6{\bf a}_i +6{\bf a}_{i+1} -4{\bf j}_i -2{\bf j}_{i+1}$$ (6.17)

$${\bf c}_i = 12{\bf a}_i -12{\bf a}_{i+1} +6{\bf j}_i +6{\bf j}_{i+1}$$ (6.18)

Substituting both expressions in Eq. 6.14 we directly find:

$${\bf v}_{i+1} = {\bf v}_i + {\textstyle\frac{1}{2}}({\bf a}_... ...t + {\textstyle\frac{1}{12}}({\bf j}_i - {\bf j}_{i+1})(dt)^2$$ (6.19)

Indeed, we have recovered Eq. 6.2, and thereby explained the mysterious factor ${\textstyle\frac{1}{12}}$ in the last term.

Let us complete our mission, by making the same derivation for the position vector, Eq. 6.1, in the Hermite scheme. Using again the snap and crackle expressions derived above, we find:

$${\bf r}_{i+1} = {\bf r}_i + {\bf v}_i dt + ({\textstyle\frac... ...}{20}}{\bf j}_i - {\textstyle\frac{1}{30}}{\bf j}_{i+1})(dt)^3$$ (6.20)

Using the same trick we employed in Eq. 6.9 to factor out the velocity terms, and using Eq. 6.19, we can rewrite the above expression as:

$${\bf r}_{i+1} = {\bf r}_i + {\textstyle\frac{1}{2}}({\bf v}_... ... + {\textstyle\frac{1}{120}}({\bf j}_i + {\bf j}_{i+1})(dt)^3$$ (6.21)

While this result still looks quite different from Eq. 6.1, we claim that it is identical up to fourth-order in $dt$, which is all we need. Our final step thus parallels the discussion following Eq. 6.9 for the leapfrog, where we had to show how terms up to second-order were identical.

First we rewrite the above equation in terms of quantities defined at $t=i$:

$${\bf r}_{i+1} = {\bf r}_i + {\textstyle\frac{1}{2}}({\bf v}_i + {\bf v}_{i+1}) dt + {\textstyle\frac{1}{10}}{\bf a}_i(dt)^2 - {\textstyle\frac{1}{10}}({\bf a}_i + {\bf j}_i dt + {\textstyle\frac{1}{2}}{\bf s}_i (dt)^2)(dt)^2$$

$$+ {\textstyle\frac{1}{120}}{\bf j}_i(dt)^3 + {\textstyle\frac{1}{120}}({\bf j}_i + {\bf s}_i dt)(dt)^3$$

$$= {\bf r}_i + {\textstyle\frac{1}{2}}({\bf v}_i + {\bf v}_{i+1}) dt - {\textstyle\frac{1}{12}}{\bf j}_i (dt)^3 - {\textstyle\frac{1}{24}} {\bf s}_i (dt)^4$$ (6.22)

Here we have left out terms containing ${\bf c}_i$, since they would be proportional to at least $(dt^5)$ and only contribute to the error noise. We can similarly write out the last term of Eq. 6.1:

$${\textstyle\frac{1}{12}}({\bf a}_i - {\bf a}_{i+1})(dt)^2 = {\textstyle\frac{1}{12}}\left({\bf a}_i(dt)^2 - {\textstyle\frac{... ...\bf a}_i + {\bf j}_i dt + {\textstyle\frac{1}{2}}{\bf s}_i (dt)^2)\right)(dt)^2$$

$$= - {\textstyle\frac{1}{12}}{\bf j}_i (dt)^3 - {\textstyle\frac{1}{24}} {\bf s}_i (dt)^4$$ (6.23)

This proves the desired result:

$${\bf r}_{i+1} = {\bf r}_i + {\textstyle\frac{1}{2}}({\bf v}_... ...t + {\textstyle\frac{1}{12}}({\bf a}_i - {\bf a}_{i+1})(dt)^2$$ (6.24)