0
$\begingroup$

x and y can be plotted for the 2D trajectory of a projectile by the following two primary equations.

$x = v t \cos \theta$

$y = vt \sin \theta - \frac{1}{2} g t^2$

$where:$

$v = initial \, velocity$

$t = time$

$\theta = launch \, angle$

$g = acceleration \, due \, to \, gravity$

Can equations be derived for the x and y trajectory of the projectile if the trajectory is altered, by a given angle at a given x,y point?

I want to completely ignore momentum, air resistance, etc.

I have found it trivial to calculate the resulting trajectory using "brute" methods when the path is reflected at a perfect horizontal or vertical angle, and it may even be trivial to reflect at 45 degree angles, such as coming into contact with a flat wall, or horizontal surface, or perfectly 45 degree surface. The problem I have is I need (for coding) a way to handle contact with a surface at any 360 degree angle.

I don't want to calculate any loss of momentum from the impact of the collision at the x,y point.

I'm not sure what the proper term is for what I'm asking for, I describe what I'm trying to do as "reflecting" or "mirroring" a trajectory about a given angle at a given point.

EDIT: I've drawn a graphic to illustrate what I'm trying to say example

The top black trajectory represents a projectile's trajectory undisturbed fired from x-3,y0.

The bottom blue trajectory represents the same projectiles trajectory if it was disturbed by a 90 degree (straight up/down, or vertical) line, which the trajectory reflects off of based on the angle of the line at the point of contact (that point is x0,y2.5.)

The gravity is the same, towards the horizontal plane, or the "ground", before and after reflection, it's not changed in any way, merely the trajectory changes direction.

This is trivial to solve for a 90 degree line, or 180 degree line, or maybe 45 degree lines, but a formula or equation which plots x and y of the trajectory after the x0,y2.5 point, no matter what angle the line is at there at x0,y2.5, is what I'm after.

$\endgroup$
  • $\begingroup$ David K, I've made an edit to the original question and added an image for explanation $\endgroup$ – Luke Allen Apr 11 '16 at 12:57
1
$\begingroup$

I would like to modify your original equations slightly by affixing subscripts to $v$ and $\theta$ to acknowledge the fact that they use the angle and speed of the projectile's motion at a particular point in space and time, namely, at the initial point $(x,y) = (x_0,y_0)$ and time $t = t_0$, and also to take into account that $x_0$, $y_0$, and $t_0$ are not necessarily all zero. (In particular, in your example in the question, $x_0 = -3$.) The result is \begin{align} x &= x_0 + v_0 (t-t_0) \cos \theta_0, \\ y &= y_0 + v_0 (t-t_0) \sin \theta_0 - \tfrac12 g (t-t_0)^2. \end{align} There are other ways to express this, but this form of the equations is relatively simple to write. (The equation measures the "extra" amount of $x$, $y$, and $t$ past the point $(x_0,y_0)$ and time $t_0$.)

We now consider another point in space and time further along the initial trajectory, at the point $(x,y) = (x_1,y_1)$ and time $t = t_1$. Note that you can uniquely determine $x_1$ and $y_1$ for any $t_1$, you can uniquely determine $t_1$ and $y_1$ for any $x_1$, and given a $y_1$ there are either two possible solutions, one possible solution, or no possible solution for $t_1$ and $x_1$. But however you decide to determine the values, we want to know all three values $x_1$, $y_1$, and $t_1$.

We also want to know the speed $v_1$ and angle $\theta_1$ of the initial trajectory as it reaches $(x_1,y_1)$. Taking the horizontal and vertical components of the velocity at time $t$ as a function of $t$, we have \begin{align} v_x(t) &= v_0 \cos \theta_0, \\ v_y(t) &= v_0 \sin \theta_0 - g(t-t_0). \end{align} The speed at time $t$ is therefore \begin{align} v(t) &= \sqrt{(v_0 \cos \theta_0)^2 + (v_0 \sin \theta_0 - g(t-t_0))^2}. \end{align} The angle of motion at time $t$ is either $$ \theta(t) = \arctan\frac{v_0 \sin \theta_0 - g(t-t_0)}{v_0 \cos \theta_0} \quad\text{or}\quad \theta(t) = \arctan\frac{v_0 \sin \theta_0 - g(t-t_0)}{v_0 \cos \theta_0} + \pi $$ depending on whether $\cos\theta_0$ is positive or negative, respectively. I'll assume from here on that $\cos\theta_0$ is positive. So at time $t = t_1$, the speed and direction of the projectile are \begin{align} v_1 &= \sqrt{(v_0 \cos \theta_0)^2 + (v_0 \sin \theta_0 - g(t_1-t_0))^2}, \tag1\\ \theta_1 &= \arctan\frac{v_0 \sin \theta_0 - g(t_1-t_0)}{v_0 \cos \theta_0}. \tag2 \end{align}

Now suppose there is a flat surface through the point $(x_1,y_1)$ at angle $\alpha$, where $\alpha$ is the angle between that surface and the positive direction of the $x$ axis; that is, $\alpha = 0$ for a horizontal surface, $\alpha=\frac\pi2$ ($90$ degrees) for a vertical surface, and $\alpha=\frac\pi4$ ($45$ degrees) for a surface at $45$ degrees from horizontal, sloping upward to the right. If we have perfect reflection of a projectile moving in the direction $\theta_1$ as it strikes that surface, the reflected path of the projectile will start at the angle $2\alpha - \theta_1$.

So what we need to do is to reapply the equations for projectile motion, but substitute $2\alpha - \theta_1$ as the new "initial angle", and translate the "initial position and time" from the point $(x_0,y_0)$ and time $t_0$ to the point $(x_1,y_1)$ and time $t_1$. We do this by substituting $x_1$ for $x_0$, $y_1$ for $y_0$, and $t_1$ for $t_0$ in the equation of motion. The result is \begin{align} x &= x_1 + v_1 (t - t_1) \cos (2\alpha - \theta_1), \\ y &= y_1 + v_1 (t - t_1) \sin (2\alpha - \theta_1) - \tfrac12 g (t- t_1)^2, \end{align} where $v_1$ and $\theta_1$ are defined by Equations $(1)$ and $(2)$ above. You can further manipulate these equations algebraically to get them into the exact form you would like.


Alternative method

Another way to derive a formula is to use a little linear algebra. We still need $x_1$, $y_1$, and $t_1$ at the point of impact on the flat surface, but we avoid explicitly computing the speed or angle of the incoming trajectory at that point.

We do still need to think about the velocity of the projectile before and after the collision, but instead of speed and direction, we use the $x$ and $y$ components of velocity, $v_x$ and $v_y$, more directly. It will help to represent the initial velocity in this form as well, that is, set \begin{align} v_{0x} &= v_0 \cos \theta_0, \\ v_{0y} &= v_0 \sin \theta_0 \end{align} so that $v_{0x}$ and $v_{0y}$ be the $x$ and $y$ components of velocity at the start of the first part of the trajectory. The equation of the trajectory can then be written \begin{align} x &= x_0 + v_{0x} (t-t_0), \\ y &= y_0 + v_{0y} (t-t_0) - \tfrac12 g (t-t_0)^2. \end{align}

The components of a velocity vector at a later time $t$ in this part of the trajectory are then $$ \begin{pmatrix} v_x(t) \\ v_y(t) \end{pmatrix} = \begin{pmatrix} v_{0x} \\ v_{0y} - g(t-t_0) \end{pmatrix}. $$ (I have put the components of the vector together in a $2\times1$ matrix here because it is a convenient format for linear algebra. Rest assured that we'll stop using this format when we get to the final answer; I am using it just for the derivation of formulas.)

The next step will be to apply a transformation matrix to reflect this vector around a line parallel to the flat surface, passing through the origin. The reason we want the line to pass through the origin is because vectors have no fixed "starting" position, only a direction and magnitude; but if you imagine the "tail" of the vector at the point $(0,0)$, then the transformation we want will leave the "tail" of the vector in place and only move the "head".

We now have two choices on how to proceed, depending on how we represent the inclination of the reflecting surface.

Using the angle of the surface. If the reflecting surface makes an angle $\alpha$ with the $x$ axis, then the reflection matrix is $$ \begin{pmatrix} \cos(2\alpha) & \sin(2\alpha) \\ \sin(2\alpha) & -\cos(2\alpha) \end{pmatrix} $$ (as shown in this answer to another question), and the reflected vector is found by multiplying the incoming vector by this matrix on the left. The resulting vector is \begin{align} \begin{pmatrix} v_{1x} \\ v_{1y} \end{pmatrix} &= \begin{pmatrix} \cos(2\alpha) & \sin(2\alpha) \\ \sin(2\alpha) & -\cos(2\alpha) \end{pmatrix} \begin{pmatrix} v_{0x} \\ v_{0y} - g(t_1-t_0) \end{pmatrix} \\ &= \begin{pmatrix} v_{0x} \cos(2\alpha) + (v_{0y} - g(t_1-t_0)) \sin(2\alpha) \\ v_{0x} \sin(2\alpha) - (v_{0y} - g(t_1-t_0)) \cos(2\alpha) \end{pmatrix} \end{align}

Writing this as a system of equations without matrices, \begin{align} v_{1x} &= v_{0x} \cos(2\alpha) + (v_{0y} - g(t_1-t_0)) \sin(2\alpha), \tag3\\ v_{1y} &= v_{0x} \sin(2\alpha) - (v_{0y} - g(t_1-t_0)) \cos(2\alpha). \tag4 \end{align}

Using the slope of the surface. If the slope of the reflecting surface is $m$, that is, the surface is parallel to the line $y = mx$, then the reflection matrix is $$ \frac{1}{1 + m^2}\begin{pmatrix} 1-m^2 & 2m \\ 2m & m^2-1 \end{pmatrix}. $$ (See this question and its answer.) The reflected vector is therefore $$ \begin{pmatrix} v_{1x} \\ v_{1y} \end{pmatrix} = \frac{1}{1 + m^2}\begin{pmatrix} 1-m^2 & 2m \\ 2m & m^2-1 \end{pmatrix} \begin{pmatrix} v_{0x} \\ v_{0y} - g(t_1-t_0) \end{pmatrix}. $$ If you do the matrix multiplication, the result is equivalent to the system of equations \begin{align} v_{1x} &= \frac{1}{1 + m^2}((1-m^2)v_{0x} + 2m(v_{0y} - g(t_1-t_0))), \tag5\\ v_{1y} &= \frac{1}{1 + m^2}(2m v_{0x} - (1-m^2)(v_{0y} - g(t_1-t_0))). \tag6 \end{align}

Of course this does not work for a vertical line of reflection, but in that case we simply change the sign of $v_x$ at the point of impact.

Assembling the final result. We already know how to write the equation of a trajectory that starts with given $x$ and $y$ velocity components from a given point at a given time; all we need to do is to take the values of $v_{1x}$ and $v_{1y}$ (either from Equations $(3)$ and $(4)$ or from Equations $(5)$ and $(6)$) and plug them into these equations: \begin{align} x &= x_1 + v_{1x} (t-t_1), \\ y &= y_1 + v_{1y} (t-t_1) - \tfrac12 g (t-t_1)^2. \end{align}

While this may seem like a long procedure, most of the writing above is just explanation and proof; to actually use the procedure, after determining $x_1$, $y_1$, and $t_1$ you merely take four of the equations above (selecting the ones that suit the way your surface is described) and start evaluating them, using the known data.

$\endgroup$
  • $\begingroup$ This seems exactly right, thank you!! $\endgroup$ – Luke Allen Apr 11 '16 at 13:14
  • $\begingroup$ Note that the "original" equations did not take into account starting the first part of the trajectory at $x=-3$ and $y=0$, so I've written that into my "original" equations now. This makes the "original" equations a little more complicated, but it also makes it easier to rewrite them to get the equations for the reflected trajectory. $\endgroup$ – David K Apr 11 '16 at 13:15
  • $\begingroup$ Yes, I understand perfectly now, this question has been on my mind for two weeks. I always thought it was possible to calculate the resultant angle after collision based on the initial angle, initial velocity, and initial position alone - without calculating the $ /theta \, 1$ and $v 1$ as you call them, but now I see. Thank you, I'm sure this will help others too, very clear manner of answering! $\endgroup$ – Luke Allen Apr 11 '16 at 13:20
  • $\begingroup$ I think you mentioned that you are a programmer. If so, you probably appreciate how useful it is sometimes to define an auxiliary variable or "helper" function that is not part of the original specification of the program. Variables such as $v_1$ and $\theta_1$ are something like that. $\endgroup$ – David K Apr 11 '16 at 13:41
  • $\begingroup$ David, the more I think about it, shouldn't it be possible to plot the x and y without calculating the impact angle and velocity with the line? The horizontal velocity doesn't change since there is only downward gravity, and in the case of colliding with a perfectly vertical line, the trajectory is mirrored by the vertical, no matter what impact angle there is. I ask out of curiosity only $\endgroup$ – Luke Allen Apr 12 '16 at 11:25
0
$\begingroup$

I am also interested in this problem. I would like to share some thoughts here.

If we treat the problem as one of reflection, then we must say which side of the mirror is silvered ( reflecting side of the mirror). For the lower blue curve (let us call it 'past')and the vertical position of mirror with left side mirrored, the top blue curve will not be the reflection(image). On the other hand, if we have the right side of the parabola ( which if call the 'future') and if the mirror is silvered on the right side then the reflection would be the top blue curve. In other words, if the object and the position of the mirror are specified, we know the shape and position of the image. The top and the bottom blue curve do not correspond to object and image in the left side silvered vertical mirror in the positions shown. So what we are seeking is not a simple reflection. Future reflected cannot give past or past reflected cannot give future. Elastic collision problem and reflection have some similarities and some dissimilarities. For example mass plays no role in reflection of light.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.