Best launch angle and velocity: HANDS DOWN..

[Symph]

STEMpunk’s shooter, optimized with applied mathematics:

Ever wonder what the single greatest, most spectacular, ABSOLUTE BEST angle and velocity to shoot your game piece at is?

Well, for us, it’s 39 degrees at 42.5 feet per second.

(From ~43.3in above the ground. Giving us a 13.5 ft thick strip of the playing field to make the shot on.)

If you want to optimize your robot like we did ours, for 2014 or any other year involving a goal and a projectile, I would encourage you to read on and find out how got these numbers.

A week into the build season our robot was well past the first basic stages of design. Our team, having decided on the layout of our drivetrain (utilizing an unusual application of the mechanum drive) and our shooter, (a virtually one piece rig functioning as an all in one, passing, shooting, ground/ air pickup mechanism) needed to know now, what results our shooter should produce to classify as a “good shooter”.
The algorithms we contrived to give us this definition, if utilized properly, should afford most teams with the capability to find that one combination of angle and velocity giving them the single greatest, most spectacular, ABSOLUTE BEST strip of the playing field to make the shot from. You can find our results above. This is how we did it:

(See "MathProofCondensed.docx" attachment.)

No doubt you will want someone who enjoys rigorous mathematics suiting these calculations to your own robot, but it’s worth the work, as this not only draws a bold line between a "bad shooter" and a "good shooter", but also marks the difference between a "good shooter" and a "perfect shooter".

[Ether]

Originally Posted by Symph

Well, for us, it’s 39 degrees at 42.5 feet per second.

That's an incomplete specification. You need to state what height above the floor that is specified at.

[Symph]

Thanks, fixed it.

[Ether]

Uh-oh. Back to the drawing board:

Y_max = ½*a*t²

t_max = sqrt(2*Y_max/a) = 0.58

[Lil' Lavery]

I disagree. This angle may provide the largest evelope to score from, but that doesn't make it hte best launch angle and velocity. There will be far superior angles in order to score from particular strategic areas of the field, or even more namely, present an easily catchable truss pass. I very much prefer a soft lob over the truss to a line drive if your intent is to get the catch points.

[Symph]

OOOOH! Never mind, I see where I went wrong! I completely neglected the fact that position was the integral of velocity. I am going to fix this, just give me a minute.

[waialua359]

The ultimate proof in my book, is how often you score and how high of a shooting percentage, WITH defense applied.

[Ether]

He's inspired by the "M" part of STEM. That's a good thing.

[bduddy]

Originally Posted by Lil' Lavery

I disagree. This angle may provide the largest evelope to score from, but that doesn't make it hte best launch angle and velocity. There will be far superior angles in order to score from particular strategic areas of the field, or even more namely, present an easily catchable truss pass. I very much prefer a soft lob over the truss to a line drive if your intent is to get the catch points.

Originally Posted by waialua359

The ultimate proof in my book, is how often you score and how high of a shooting percentage, WITH defense applied.

^ And therein lies the difference between the 'M' and the 'E' in STEM!

[Symph]

I have revised the entire proof, sorry about the inconvenience, it should be sound now. However, considering the fact that I was thinking precisely that same thing last time I submitted this, I would encourage you to feel free to check my math again!
What say you, Ether?

[Ether]

Originally Posted by bduddy

^ And therein lies the difference between the 'M' and the 'E' in STEM!

FRC is a big tent. There's room for everybody.

[Ether]

Originally Posted by Symph

I have revised the entire proof, sorry about the inconvenience, it should be sound now. However, considering the fact that I was thinking precisely that same thing last time I submitted this, I would encourage you to feel free to check my math again!
What say you, Ether?

You forgot to edit your original post. It still says "42.5 feet per second" and "13.5" ft scoring range.

[Ether]

Originally Posted by Symph

I have revised the entire proof, sorry about the inconvenience, it should be sound now. However, considering the fact that I was thinking precisely that same thing last time I submitted this, I would encourage you to feel free to check my math again!
What say you, Ether?

This time around I didn't review all your steps again. Instead I did the following two things:

Thing1:

I took your "final answer" of V=29.53 f/s, angle=39 degrees, launch height = 3.61042 feet and calculated the trajectory from that. Your apex occurs at x=13.2555 ft and Rs is 11.3 ft.

Here's the equation for the trajectory using 29.53 f/s & 39 degrees:

y = -0.0305452*x^2 + 0.809784*x + 3.6104

(x is horizontal distance from launch point and y is height above floor)

Thing2:

I did my own calculation to find launch speed and angle to get the desired Rs=13.5 feet.

The result is

launch speed = 31.970 ft/sec at launch angle 35.55 degrees

The trajectory using those numbers is

y = -0.0237769*x^2 + 0.714568*x + 3.6104

max height above floor = 8.9792ft @ 15.0265ft from launch point

Bottom line: whatever you're doing is certainly close enough engineering-wise for FRC application. Mathematically, I suspect you are getting some round-off error due to the calculation method you used.

PS: You forgot to edit your original post. It still says "42.5 feet per second" and "13.5" ft scoring range.

[Uriah]

Originally Posted by Ether

He's inspired by the "M" part of STEM. That's a good thing.

Ether,

Not only is he inspired by it, he is driven by it. He's a 16 year old homeschool kid that taught himself calculus. He's a very bright young man and STEMpunk is lucky to have him for another year!

[Symph]

Ether, I am sorry to say that we are both unfortunately confused. You see, I had redone the math and was going through the process of updating my post in stages; Replacing my "MathProofCondensed.docx" attachment, replacing the graph for my trajectory in the original post and finally responding to you, when my Wi-Fi cut out for a full day. Which I thought was fine, considering that I had finished updating, until I realized that I had neglected to explain why I still went for 39 degrees even with an alternate velocity, and left my original sentence "Well, for us, it’s 39 degrees at 42.5 feet per second." unchanged.
I will first begin by explaining that I chose 39 over 36 degrees knowing full well that it would result in an Rs of 11.5ft rather than a full 162in or 13.5ft, because although 36 degrees does give us the greatest strip of the field, it's furthest point lies in the middle zone rather than the beginning of the first, where we will not be shooting from leaving a potion of that strip wasted, and it's closest point is several feet away from the goalie zone, forcing us to shoot from further away and over build our robot.
If you graph the function "R''s=-115.7137 *csc(⁡a)*cot(⁡a)" in my proof, you will see that one of it's many zeros is in indeed ~36 degrees, and does indeed, as you calculated, result in Rs= 13.5ft.
As you can see now, my math is right, but my final decision does not match what I first defined as "optimal", and so in the name of peace, harmony and all things good in the world, I am going to update my post, once more, this time sticking to my original postulate of "best angle and velocity => biggest Rs".
Hopefully this time, I will have it right,
Wish me luck Ether!