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# What Equation Draws This Graph? (and can you do better?)

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Having not done much of this since highschool, i'm appealing to you math gurus for this one. Google can only teach me so much. I need a real person. For a math model in my game, i need to produce this graph (or just the upper right quadrant of it anyway): I need an equation for this with a few gotchas. Context: It's meant to represent diminishing returns. This is a general enough statement to get the problem communicated without bothering you with all the details. But let's call the Y axis "Percentage Effectiveness" and the X axis "Distance". That is: the further away you get, the worse the effectiveness is. Assume that the blue dot is the "start" and that it is always y=100 (to represent "100% effective"). The red dot is the "end" at which you have 0% effect. I want the graph to ride along pretty straight for a long while then drop off sharply at the end. This is what i came up with to produce the above graph, just goofing off on a graphing calc online: y = 100 - ( (0.05*x)^4 ) Here is my gotcha: The endpoint on the slope (rightside x-intercept) needs to be a specific number (call it "Z") and the graph needs to adjust accordingly. That is to say, "You achieve 0% Effect at a Distance of Z." I hope i explained that right. What i want to know is: - Is there a simpler formula for this or am i making it too complicated? - How can i keep the same basic graph but "stretch" it horizontally to hit a specific x-intercept? Thanks for your help. I really appreciate it.

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I thought that looked like a quartic. Looks like I still remember that stuff!

Anyway, here's my opinion on your problem:
Quote:
 Original post by leiavoia- Is there a simpler formula for this or am i making it too complicated?

Your formula is quite simple as it is, at least to me. You could go with a quadratic or something inversely exponential or logarithmic if you want a different curve, but if the quartic model fits your needs best, then go with it.

Quote:
 - How can i keep the same basic graph but "stretch" it horizontally to hit a specific x-intercept?

It's the coefficient in front of the x term in your equation that controls the "stretchiness" in the direction of the x-axis. So in your example, it's the 0.05. Halve that, and the graph will dilate out by a factor of 2.

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y = 100 - ( (0.05*x)^4 )

You want to adjust the 0.05 so that your graph is 0 at a certain X.

Let's call it a and solve.

0 = 100 - ( (a*x) ^ 4)
(a*x) ^ 4 = 100
a*x = 100^(1/4)
a = 3.1622/x

That's all you need, unless I've made some stupid mistake somewhere.

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If your function has the value you want at input X, and you want it to have that same value at input Y, simply multiply the X by Y/X:

  y = 100 - ( (0.05*(x*(X/Y))^4 )

More generally, if you want value V at point Q, you can analytically solve this through finding derivative = 0 of the function (F - (V*x/Q)).

When it comes to diminishing returns, my favorite is the atan function. You can just keep going and going and it never QUITE gets to 1.0 :-)

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You guys are quick. Thanks!

I guess my original equation was fairly decent then. I've got a good model in mind now with some numbers to go with it. Thanks again!

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Quote:
 Original post by leiavoiaContext: It's meant to represent diminishing returns. This is a general enough statement to get the problem communicated without bothering you with all the details. But let's call the Y axis "Percentage Effectiveness" and the X axis "Distance". That is: the further away you get, the worse the effectiveness is.

That's not really a "diminishing returns" graph. Such a graph would be tending to zero, not decreasing at an increasing rate. This one is more like an application of an inverse-squared law (e.g. radiance y at a distance x) or something similar.

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You are right, but it fits my model better. I'm actually modeling "planetary habitability" for space-faring races in a strategy game i'm designing. I want races to be able to settle planets in their general "comfort zone," with diminishing returns the further away from their zone that they get. However, there needs to be a point at which a planet is totally inhospitable even to technologically advanced life (that is, y=0).

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Quote:
 Original post by leiavoiaYou are right, but it fits my model better. I'm actually modeling "planetary habitability" for space-faring races in a strategy game i'm designing. I want races to be able to settle planets in their general "comfort zone," with diminishing returns the further away from their zone that they get. However, there needs to be a point at which a planet is totally inhospitable even to technologically advanced life (that is, y=0).

The answers you've received so far look great to me, but I thought I'd just throw this one in the pot for flavour [smile] You could try one-sized Gaussian, which is more in keeping with what kSqaured suggested. f(x) = exp(-a*x^2), where a > 0. The shortcoming of this is that there is no value of x where f(x) = 0, if that is a problem. However, if you scaled f(x) properly, you will have a nice probabilistic interpretation for 'normal' planetary conditions.

-Josh

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From "Texturing & Modeling" second edition, by Ebert et al. there is an equation that *may* be interesting to you - I use it all the time.

It seems you want the slope at x=0 to be 0, good. For diminishing returns, we would like the slope at x=max to also be 0. The equation that works for this can be found in chapter 2 page 26-27, called a "smoothstep."

y=3x^2 - 2x^3 is zero at x=0 and 1 at x=1.

Invert:
y=1 - (3x^2 - 2x^3), now 1 at x=0 and 0 at x=1.

normalize your range first before putting it into the equation:
a = minx, b = maxx
x = (x - a) / (b - a)

Increase the range by r:
y = r * (1 - (3((x-a)/(b-a))^2 - 2((x-a)/(b-a))^3))

So, there is an alternative. Again, not an accurate model of diminishing returns, but maybe better than a vertical cutoff of the x-axis at x=max.

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Then, there is also the cosine function (slopes 0 at minx and maxx).
y=cosx gives 1 at x=0 and -1 at x=pi

using same adjustment calculations as my previous post, we have:

adjust to domain [0,1]
y = cos(pi * x), yields 1 at x=0 and -1 at x=1

adjust range to [0,1]
y = (cos(pi * x) + 1) / 2, yields 1 at x=0, 0 at x=1

adjust range to [0, r]
y = r * (cos(pi * x) + 1) / 2

adjust domain to [a, b]
y = r * (cos(pi * ((x - a) / (b - a))) + 1) / 2
This last one is r at x=a and 0 at x=b

These techniques for adjusting domain and range can be applied to most any equation. Good luck.

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