Projectile Motion

[MechEng83]

In the numerous threads, I've seen statements that shows me a lot of posters only know enough about projectile motion to be dangerous. I hope to clarify a few points, which should be applicable regardless of how you launch the game piece. Caveat: this discussion will neglect air resistance.

False: The ball will travel in a straight line before it falls.
Truth: The path of a projectile is a parabola.
Clarification: If your speed is fast enough, you can approximate the path as a straight line until the gravitational acceleration becomes significant.

False: 45 degrees is the "best" launch angle to get over the truss/in the high goal.
Truth: Maximum horizontal distance when the ball returns to its starting height is achieved when the projectile is launched at 45 degrees.
Clarification: Assuming a fixed launch speed, a steeper launch angle will cause the ball to travel higher, which means it's in the air for a longer period of time. When launching from a height above zero, an angle less than 45 degrees gives the furthest horizontal distance. (For those of you more advanced in mathematics, solve for the slope of the parabola at the point it hits the ground to be -1)

[AdamHeard]

Originally Posted by MechEng83

In the numerous threads, I've seen statements that shows me a lot of posters only know enough about projectile motion to be dangerous. I hope to clarify a few points, which should be applicable regardless of how you launch the game piece. Caveat: this discussion will neglect air resistance.

False: The ball will travel in a straight line before it falls.
Truth: The path of a projectile is a parabola.
Clarification: If your speed is fast enough, you can approximate the path as a straight line until the gravitational acceleration becomes significant.

False: 45 degrees is the "best" launch angle to get over the truss/in the high goal.
Truth: Maximum horizontal distance when the ball returns to its starting height is achieved when the projectile is launched at 45 degrees.
Clarification: Assuming a fixed launch speed, a steeper launch angle will cause the ball to travel higher, which means it's in the air for a longer period of time. When launching from a height above zero, an angle less than 45 degrees gives the furthest horizontal distance. (For those of you more advanced in mathematics, solve for the slope of the parabola at the point it hits the ground to be -1)

I wouldn't make generalizations like this without also commenting that drag significantly affects the trajectory.

[puneeth.meruva]

I would assume that you can't exactly ignore air resistance with such a large projectile.

[MechEng83]

I don't really want to argue semantics, but a projectile is a body for which the only force acting upon it is gravity. Ergo, the drag specifically makes it NOT a projectile motion problem.

Some of the technical posts in this thread: http://www.chiefdelphi.com/forums/sh...d.php?t=124222 might be helpful to those who would like to model more of the physics. As noted in that thread, whether or not drag is significant is a judgement call and very much depends on the velocities involved.

[Bob Steele]

Originally Posted by MechEng83

When launching from a height above zero, an angle less than 45 degrees gives the furthest horizontal distance. )

Not true
It gives you a greater horizonatal VELOOCITY

Horizontal Distance travelled is a combination of the horizontal velocity and the time it is applied.

The time that is applied is a function of how high the ball travels ...


You have to always consider BOTH the x and the y components of the velocity in order to determine the horizontal distance travelled.

In fact the maximum horizontal range is at 45 degrees.

In this problem (excluding air resistance) you have to solve for the height you need at the distance you want to shoot from. The solution to shooting has infinite combinations of velocity and angle and distance to the goal that are solutions.

Of course there are some velocities and angles and distances that would not provide a solution. IE launching in the opposite direction, launching straight up, Launching with a velocity of 0 feet per second...etc etc

[PerpetualMotion]

Originally Posted by Bob Steele

Not true
In fact the maximum horizontal range is at 45 degrees.

45 deg gives maximum horizontal range when your starting y position and finishing y position is the same

Originally Posted by MechEng83

When launching from a height above zero, an angle less than 45 degrees gives the furthest horizontal distance.

In the case of Aerial Assist, the ball is being launched from a distance below zero, since the high goal is higher than the maximum height of a robot. Therefore, in the context of this game, the maximum distance for a given velocity will be greater than 45 degrees (assuming no air resistance)

[MechEng83]

Originally Posted by PerpetualMotion

45 deg gives maximum horizontal range when your starting y position and finishing y position is the same



In the case of Aerial Assist, the ball is being launched from a distance below zero, since the high goal is higher than the maximum height of a robot. Therefore, in the context of this game, the maximum distance for a given velocity will be greater than 45 degrees (assuming no air resistance)

I should have clarified. The solution I gave was for the ball landing below the launch point. The point I wished to make was that the maximum distance needs the destination to be at 45 degrees, and your vertical position relative to that will affect the angle which gives maximum launch velocity. 45 degrees is solution to the special case where h(0) = h(final).

[Ether]

Parabola Equations:

http://www.chiefdelphi.com/forums/sh...d.php?t=124392

[bEdhEd]

The best angle is 45 degrees directly at the floor. BOOM

[cmrnpizzo14]

False: Math like this always generates the correct design for your robot.
Truth: The math is helpful but in the end I'm just going to prototype a few designs and see which one works best.



303rd post goes out to the TEST Team, awesome job guys.