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Vector Described


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#1 brycelarson

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Posted 29 July 2009 - 02:59 PM

OK, since it's a common term used constantly by CockerPunk and me - I thought I would whip up a quick tutorial on what the term vector is and what we think it means.

I'll use the following data - it's from a previous test:

Posted Image

And here's a graph of that data - the blue square is the mean impact location - or the center of this group.
Posted Image

Here's the data corrected to 0,0. You don't need to do this to calculate the vector - but it will make this tutorial make more sense.
Posted Image

Here's the data again - the Standard Deviations of X and Y have been graphed - as the green triangle.
Posted Image

Here's the hypotenuse of the SD - OR, a line from 0,0 to the point where the X and Y standard deviations land
Posted Image

Here's the vector displayed as a radius of a circle. In theory that circle should encompass 68% of the shots. In this case it contains 15 of the 22 shots - which, happily, is 68% of them.
Posted Image

Here's a graph with a circle with a radius of 2x the vector. Theoretically, this circle should encompass 95% of all shots ever fired by this setup. We have none outside - but that's because we don't have a large enough sample.
Posted Image

hope this helps.

Edited by brycelarson, 29 July 2009 - 03:00 PM.


#2 UV Halo

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Posted 29 July 2009 - 09:37 PM

Outstanding work! Based on our previous discussions, I understood it already but, this should help anyone who didn't get it!

EDIT: I strongly recommend a stickied Lexicon, and this should be in it (editable only by PunkWorks staff)

Edited by UV Halo, 29 July 2009 - 09:38 PM.


#3 chewiestmonkey

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Posted 30 July 2009 - 12:22 PM

excellent!
my understanding on this subject was somewhat sketchy, this cleared it up though!
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#4 rntlee

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Posted 30 July 2009 - 06:37 PM

ty sir, very well done.

#5 awesomo12000

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Posted 30 July 2009 - 08:52 PM

Posted Image

Sorry for my unprofessionalism...

Edited by awesomo12000, 30 July 2009 - 09:03 PM.

Posted Image



#6 brycelarson

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Posted 02 August 2009 - 08:01 AM

your 'vector' is the hypotenuse of the triangle formed by the average (mean), origin, and x/y axis (doesn't matter)

It's a good value for approximating the area of a shot grouping.


not quite - it's the hypotenuse of the triangle formed by the origin (0,0) and the standard deviation of the data in the X axis and the Y axis. standard deviation is already incorporating the mean in it's calculation - so you don't need to apply it again.

In this case I shifted the group to center the mean on 0,0 just to make the graphics more clear. that step isn't necessary to find the vector of a group.

and yes, it is a as good as we've found for comparing shot grouping and accuracy.

#7 Egomaniacal

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Posted 01 September 2009 - 06:53 PM

That doesn't make any sense to me. Is your vector just sqrt(sigma(x)^2 + sigma(y)^2)? Because that is in fact the magnitude of the vector from the mean to sigma(x) and sigma(y).

Edit: By the way you're also assuming a normal distribution, which does seem to describe the data well.

Edited by Egomaniacal, 01 September 2009 - 07:05 PM.

eiπ = − 1

#8 cockerpunk

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Posted 01 September 2009 - 08:39 PM

That doesn't make any sense to me. Is your vector just sqrt(sigma(x)^2 + sigma(y)^2)? Because that is in fact the magnitude of the vector from the mean to sigma(x) and sigma(y).

Edit: By the way you're also assuming a normal distribution, which does seem to describe the data well.


you got it!
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#9 GaryD

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Posted 23 May 2010 - 11:13 AM

Thank you for posting the information on how the vector is calculated. I just recently found this site. I have been very slowly devouring the wonderful experimental information that you have been gathering, and also had been wondering how you arrived at the vector. I find all the information you have been posting exciting. The reason is that eons ago I put together a hardcore paintball physics site. The site still laments the lack of experimental data needed to validate and verify even some of the physics. You all are doing a great job and I hope you can sustain the effort.

My specific question is related to the position on the graph axis for the "green triangle". The value plotted does not seem to correspond to the std dev of the x and y points you derived. The relative position looks to be about right. Am I missing something?

#10 brycelarson

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Posted 23 May 2010 - 03:29 PM

I'm not at the computer that I created the graph on - but I see that. There are a couple of things that might be going on. I'm guessing I typed the 5.88 into the graphing software wrong. The second thing is that when I overlayed the polar graph onto the graph of the points I seem to remember something odd going on with the software. I might have had to fudge the input numbers to make the relationship work right.

You're exactly right - that green triangle looks to be at about 8, 7.

hopefully the explanation is still helpful even with the error.

#11 brycelarson

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Posted 23 May 2010 - 03:32 PM

oh, and is this you?

http://home.comcast.net/~dyrgcmn/

you going to make it to Living Legends III next weekend?

Edited by brycelarson, 23 May 2010 - 03:32 PM.


#12 GaryD

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Posted 25 May 2010 - 06:11 PM

oh, and is this you?

http://home.comcast.net/~dyrgcmn/

you going to make it to Living Legends III next weekend?



Yes, that is me. No, I will not be at the Living Legends. Unfortunately, I have not been actively going out to play paintball the last few years. It is not that I lost my enthusiasm, but to some extent I have begun to act my age and the fact that my son is now living in Kansis City rather than Chicago had a lot to do with my playing dropping off. In addition, work began to get in the way of some of the finer things in life, like playing paintball.

#13 ChloricName

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Posted 03 April 2014 - 09:04 PM

I know this is a necro but, can anyone simplify this for a twelve year old? I want to do this on my science project. Is this important for a seventh grade experiment?

#14 UV Halo

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Posted 04 April 2014 - 01:33 PM

I'm sorry I but, I don't know where you are in terms of Math development (what are they teaching in 7th grade these days?) but, I can tell you if you dont have the skills you soon will.  To use this method you need to be able to:

 

  • Perform Addition and subtraction- I'm fairly certain you can do that.
  • Calculate the average (aka "Mean") for a group of numbers, or  know how to perform this function in a spreadsheet (i.e. Microsoft Excell, OpenOffice Calc, etc).  I think you can do this.
  • Solve for the length of a third side (Hypotenuse) of a triangle when you're given the lengths of the two other sides-  I'm not sure about this.
  • Calculate the Standard Deviation for a set of numbers (or use a spreadsheet function).  I don't think you've learned this yet.

Fortunately, there are plenty of online resources to learn all of these things (either manually or via a spreadsheet).  The question is, do you have the time in order make the deadline?  If you learn these things it'll be pretty easy to walk you through this process.

 

 






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