Well then...
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Well then...
Wed Apr 2, 2008 3:17:08 PMOk...so here's what I figured out. It's not a monumental discovery, but it's certainly not been introduced. I've searched online and haven't found anything like it...so here it goes.
I developed a formula involved in creating a graph by hand that makes plotting points, maximizing the use of your window, or developing a good interval easier.
-What? Graphing is already easy Kyle! You're useless!
Well what happens when you're graphing values like .5338267 or fractions like (61/53) on an axis with an interval that counts by another ugly decimal like .43?
-That's stupid, why would I use dumb numbers like that?
Not every graph you make is going to have pretty, round numbers. It's not uncommon that I find myself struggling to find the appropriate place on a graph for a certain point I'm trying to plot, so I'm guessing that other people have this problem too. Also, this equation I'm going to show you has other equations that can be derived from it, allowing you to maximize your window or develop a useful interval to count by.
So here's the set-up:
You have a value you want to graph, we'll call this (A). We'll assign (A) some value. How about 1.367.
You have some interval on some axis, we'll say that we're counting by .5328 on the x-axis. We'll label this interval (X)
There is then some sub-interval between the interval, or the number of boxes between each interval. This is going to be called (Z). We'll say that (Z) is 3 boxes long.
Now, who would easily be able to find the accurate location of this point on the x-axis in a timely fashion? Not many people, I'd assume.
So, here's what I came up with:
Take (A), we'll just say that (A) is some measurement of time in seconds. So we have 1.367 seconds.
Now divide (A) by the interval you're using, (X). (X) increases .5328 seconds every 3 boxes, also known as (Z).
So we have:
(A/X) or (1.367sec./0.5328sec.)
This value has no unit now, because seconds over seconds cancels out. We're left with 2.565690691 with no units.
Now take this value and multiply by (Z), number of boxes.
2.565690691 * 3 boxes = 7.697 boxes
You know know the precise location of point (A) on the x-axis with some interval (X) and some sub-interval (Z).
Now some people that are educated might argue that there doesn't have to be a sub-interval, in which case you're right. If you're going to mark every value on the x-axis then your sub-interval is going to be 1 and the original interval is going to be smaller. So if we took our (X) from the example above and divided it by 3 (Z) we could eliminate the sub-interval. But typically, people don't mark every value on the x-axis, and instead skip boxes, creating a larger interval and sub-interval.
For those that don't follow my words, here's a pretty picture from MS Paint:
**Note: Counting the boxes will actually make the graph inaccurate, instead count each line of demarcation from the origin.
I developed a formula involved in creating a graph by hand that makes plotting points, maximizing the use of your window, or developing a good interval easier.
-What? Graphing is already easy Kyle! You're useless!
Well what happens when you're graphing values like .5338267 or fractions like (61/53) on an axis with an interval that counts by another ugly decimal like .43?
-That's stupid, why would I use dumb numbers like that?
Not every graph you make is going to have pretty, round numbers. It's not uncommon that I find myself struggling to find the appropriate place on a graph for a certain point I'm trying to plot, so I'm guessing that other people have this problem too. Also, this equation I'm going to show you has other equations that can be derived from it, allowing you to maximize your window or develop a useful interval to count by.
So here's the set-up:
You have a value you want to graph, we'll call this (A). We'll assign (A) some value. How about 1.367.
You have some interval on some axis, we'll say that we're counting by .5328 on the x-axis. We'll label this interval (X)
There is then some sub-interval between the interval, or the number of boxes between each interval. This is going to be called (Z). We'll say that (Z) is 3 boxes long.
Now, who would easily be able to find the accurate location of this point on the x-axis in a timely fashion? Not many people, I'd assume.
So, here's what I came up with:
Take (A), we'll just say that (A) is some measurement of time in seconds. So we have 1.367 seconds.
Now divide (A) by the interval you're using, (X). (X) increases .5328 seconds every 3 boxes, also known as (Z).
So we have:
(A/X) or (1.367sec./0.5328sec.)
This value has no unit now, because seconds over seconds cancels out. We're left with 2.565690691 with no units.
Now take this value and multiply by (Z), number of boxes.
2.565690691 * 3 boxes = 7.697 boxes
You know know the precise location of point (A) on the x-axis with some interval (X) and some sub-interval (Z).
Now some people that are educated might argue that there doesn't have to be a sub-interval, in which case you're right. If you're going to mark every value on the x-axis then your sub-interval is going to be 1 and the original interval is going to be smaller. So if we took our (X) from the example above and divided it by 3 (Z) we could eliminate the sub-interval. But typically, people don't mark every value on the x-axis, and instead skip boxes, creating a larger interval and sub-interval.
For those that don't follow my words, here's a pretty picture from MS Paint:
**Note: Counting the boxes will actually make the graph inaccurate, instead count each line of demarcation from the origin.
Dude Lightning
2014: Year of change.
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Joined: December 5, 2007
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2014: Year of change.
Joined: August 3, 2001
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Posts: 10753
Rep:
Joined: December 5, 2007
Status: Offline
Posts: 359
Rep:
Joined: December 5, 2007
Status: Offline
Posts: 359
Rep:
2014: Year of change.
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Joined: December 5, 2007
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