User:Neil Parker/Quaternions in Projectile Motion

Introduction
In a previous Physics Forums article entitled “How to Master Projectile Motion Without Quadratics”, PF user @Kuruman brought to our attention the vector equation  $$\frac{|V_0 \times V_f|}{g} = R$$ and lamented the fact that:

“Equally unused, untaught and apparently not even assigned as a “show that” exercise is Equation (4) that identifies the range as the magnitude of the cross product of the initial and final velocity divided by g.”

In this article, we reproduce and make use of  equations presented in a 60s vintage article from the American Journal of Physics entitled “Convenient Equations for Projectile Motion”. The article abstract indicates:

"Quaternion multiplication of the basic vector equations for uniformly accelerated motion gives 2 equations: $$v^2 = u^2 + 2\vec{a}.\vec{s}$$ and $$\vec{a} \times \vec{s} = \vec{v} \times \vec{u}$$ which provide a solution for some projectile problems."

It will be noted that the first equation is essentially an expression for conservation of mechanical energy whilst the second is exactly that referred to by PF user @Kuruman above. Whilst - with the studied understatement typical of  60s vintage AJP articles such as this one  - we are told that the equations “provide a solution for some projectile problems”, we intend to demonstrate that virtually every single projectile motion problem we have found (in the annals of Physics Forums homework questions and elsewhere) can be solved using these two equations in tandem. Echoing the comment of PF user Kuruman, J. Gibson Winans – author of the quaternions article – writes:

"Equation (7) (ie $$\vec{a} \times \vec{s} = \vec{v} \times \vec{u}$$) is one which seems to have had little use in describing projectile motion."

We hope to further demonstrate that PF user @Kuruman's lament is entirely justified and that  J. Gibson Winan's statement above is just one step too far in modesty even given the  AJP penchant for  ‘studied understatement’!

Quaternion Derivation
We begin with a disclaimer: the current author has no knowledge of quaternions other than that researched for this article so the derivation which follows is quoted ‘verbatim’ from the article by J. Gibson Winans. He begins by presenting the problem of a projectile fired at one elevation being required to hit a target at a different elevation:

"To solve this problem conveniently,  it suffices to treat the basic equations of accelerated motion as vector equations and derive additional equations from them. Writing

and

gives the two basic vector equations in a  convenient form. Elimination of v between (1)  and (2) gives the much-used vector equation:

To obtain two additional equations to assist in the solution of this problem, Eq. (2) can be multiplied by Eq. (1) to obtain

Author's note: Variable t is eliminated when the above product is obtained. However, it can easily be retrieved either from Eqn 1 or from Eqn 2 if required.

"Equation (4) contains three products of vectors. These are neither scalar products nor vector products, but are quaternion products. The quaternion product of two vectors is the scalar product plus the vector product. Equation (4) therefore is equivalent, after cancelling, to

Equation (5) can be separated into two equations, one a scalar equation and one a vector

Since: $$ \text{-} \vec{u} \times \vec{v}=\vec{v} \times \vec{u}$$ and the factor 2 cancels on both sides of equation ,the vector equation is:

Equations (1) to (7) above are reproduced in full from American Journal of Physics 29, 623 (1961); https://doi.org/10.1119/1.1937861 with the kind permission of the American Association of Physics Teachers.

In the case of projectile motion, Equation 6 takes the well known form $$v^2-u^2=2gh$$ whilst equation 7 can be written as $$\vec{g} \times \vec{R} = \vec{v} \times \vec{u}$$ where R is the horizontal range of the projectile.

We will illustrate the technique of using these 2 equations in tandem by providing 2 worked examples: one numeric and one symbolic. There are many other examples one could choose from, so we just provide a couple of 'aperitifs'. We invite our readers to try out the technique on any selected 2D projectile problem and share their experience in the comments section.

Worked Example 1
The numeric problem to be solved is to obtain the horizontal range and hence the time of flight for a rock hurled from a volcano at an angle of $$35^{\circ}$$ to the horizontal at a velocity of 25m/s landing at a point 20m below its starting point. We will make use of the vector facilities available from Wolfram Alpha. Note that Equation 7 will be submitted in vector form whilst Equation 6 is first employed to determine $$v_y$$. R (horizontal range) is the variable being sought and - once obtained - can simply be divided by $$v_{x}=25\cos(35^{\circ})$$ to obtain time of flight.

With reference to the Wolfram Alpha entry below, the input data is as follows:


 * $$\vec{g}=(0,-9.8)$$ $$\vec{R}=(R,0)$$ $$\vec{u}=(25cos(35^{\circ}),25sin(35^{\circ})).$$

The horizontal component of  velocity $$u_x=v_x=25cos(35^{\circ})$$ does not change but we will employ Equation 6 to calculate $$v_y$$:


 * $$v_y=\sqrt{(25sin(35))^2+2\times9.8\times20)}.$$

Hence the overall vector entry for $$\vec{v}$$ will be:
 * $$\vec{v}=\left(25cos(35^{\circ}),-\sqrt{(25sin(35))^2+2\times9.8\times20)}\right).$$



Dividing by $$v_x=25\cos(35^{\circ})$$ we obtain $$t=3.958\;s.$$ for the time of flight. The reader will find a 'conventional' solution to the same problem here. . (you might have to scroll down somewhat.)

Worked Example 2: Range up a Slope
This will be a symbolic problem whereby we derive an equation for the range w of a projectile fired up a slope of angle $$\alpha$$ at a velocity v and angle $$\theta$$ from the horizontal (clearly $$\theta>\alpha$$). Preparing our entries for submitting and solving the vector equation on Wolfram Alpha, we write:


 * $$\vec{g}=(0,-g)$$
 * $$\vec{R}=(w\cos\alpha,0)$$
 * $$\vec{u}=(u\cos\theta,u\sin\theta).$$

Once again the horizontal component of velocity $$u_x=v_x=u\cos\theta$$ does not change but we will need equation 6 to obtain an expression for $$v_y$$.For this we write the vertical height gained by the projectile as $$w\sin\alpha$$. Then:


 * $$v_y=-\sqrt{((u\sin\theta)^2-2gw\sin\alpha)}$$.

Hence the overall vector entry for $$\vec{v}$$ will be:


 * $$\vec{v}=\left(u\cos\theta,-\sqrt{((u\sin\theta)^2-2gw\sin\alpha)}\right).$$

We may now proceed with submission of the vector equation on Wolfram Alpha.

At first glance, the solution appears to be all negative, but the term $$-u^2\sin(\alpha-2\theta)$$ can be written as $$+u^2\sin(2\theta-\alpha)$$ and the reader may verify this as the correct expression for range up a slope here.

It is true that Wolfram Alpha "chomps" through a lot of algebra to reach the above solution but if the reader cares to examine the 'step-by-step' solution, it will be clear that there is nothing particularly complicated about the process. Wolfram Alpha simply gives us the assurance that there won't be any mistakes in the algebra leading to the range equation above. And that we can have every confidence in employing the vector equation as indicated since the results are exactly the same as those obtained by more 'conventional' means.

It is worth noting that determination of the range down a slope (rather than up) entails nothing more than a single sign change in the above vector equation:

.

Alternate Derivations
We do not have to use quaternions in order to derive either the dot product or cross product equations. We have already mentioned (for example) that the dot product equation reflects conservation of mechanical energy:


 * $$\text{½}mv^2 = \text{½}mu^2 + F \Delta s \Rightarrow \text{½}mv^2 = \text{½}mu^2 + ma \Delta s  \Rightarrow v^2 - u^2 = 2 a \Delta s $$

As presented in a related insights article - we can use Calculus:


 * $$a_s=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v_s\frac{dv_s}{ds}\implies v_s dv_s=a_s ds$$


 * $$\int_{v_{0s}}^{v_s}v_s dv_s=a_s\int_{s_0}^s ds \implies v_s^2-v_{0s}^2=2a_s(s-s_0)=2a_s\Delta s.$$

A technique which parallels the quaternion approach presented above (but confined to one dimension) is to re-write basic SUVAT equations in the following forms:


 * $$v = u + at \Rightarrow v-u=at$$
 * $$ \frac{v+u}{2}t=\Delta s \Rightarrow v+u=\frac{2 \Delta s}{t}$$

These two equations can be multiplied eliminating t as a parameter and yielding the 'dot product' part of the quaternion equation:


 * $$v^2-u^2=2a\Delta s$$

The cross product part of the quaternion equation can be derived as follows:


 * $$\vec{a}\times\vec{s}=\vec{a} \times (\vec{u} t+\text{½} \vec{a} t^2)=(\vec{a}t )\times \vec {u} =(\vec{v}-\vec{u})\times \vec{u} =\vec{v}\times\vec{u}.$$

Summary and Conclusion
This article has drawn extensively on J. Gibson Winans's article describing the quaternion-based derivation of two equations that can be used in tandem to solve projectile problems. We have provided two demonstrations of the use thereof in the worked examples above and we hope that these 'aperitifs' will whet the reader's appetite to try out the technique on other projectile problem(s) which may arise either in PF's 'homework help' forums or from elsewhere. As mentioned earlier, readers are encouraged to share their experiences with us via the comment section for this article.