Trying to "create" an equation

Spike Spiegel

New member
Joined
Feb 12, 2009
Messages
168
Reaction score
0
Points
0
Trying to "create" an equation. And failing miserably.

I've been working on a science fiction story for a ridiculously long amount of time, and I'm trying to nail down the properties of a particular piece of fictional technology. I'd like to do this as accurately as I can, to the point where an equation or two can describe it.

I'm trying to show that as the distance from a central point increases, the probability of accurately targeting a second point decreases. Naturally I looked at the inverse-square law. Unfortunately, I'm not too great with math beyond basic algebra and geometry.

What I want is for the probability to be 100% (or 1.00) when the target point is the same as the central point (radius = 0), then decrease from there, somehow according to the inverse-square law. I also want to be able to show that some other factor can multiply this probability (increasing the range at which the second point can be reliably targeted).

I'm not asking anyone to do the work for me, but I described it so that someone can hopefully point me in the right direction. Can one of you resident geniuses give me some tips, or tell me what I should be studying so I can figure this out?
 

Izack

Non sequitur
Addon Developer
Joined
Feb 4, 2010
Messages
6,666
Reaction score
3
Points
0
Location
The Wilderness, N.B.
Instead of looking at equalities, look at proportions.

Let r be distance from the first point, and a be accuracy.

a is inversely proportional to the square of r:

a ~ r^(-2) or a ~ 1 / r^2

You want accuracy to be directly proportional to some other value. Let that be x:

a ~ x

Put those together in an equation:

a = x / r^2


Unrelated: I noticed some orbinauts have used "math" tags to pretty up their equations. Can I find an explanation of this anywhere?
 

martins

Orbiter Founder
Orbiter Founder
Joined
Mar 31, 2008
Messages
2,349
Reaction score
17
Points
0
Website
orbit.medphys.ucl.ac.uk
Trying to "create" an equation. And failing miserably.

I've been working on a science fiction story for a ridiculously long amount of time, and I'm trying to nail down the properties of a particular piece of fictional technology. I'd like to do this as accurately as I can, to the point where an equation or two can describe it.

I'm trying to show that as the distance from a central point increases, the probability of accurately targeting a second point decreases. Naturally I looked at the inverse-square law. Unfortunately, I'm not too great with math beyond basic algebra and geometry.

What I want is for the probability to be 100% (or 1.00) when the target point is the same as the central point (radius = 0), then decrease from there, somehow according to the inverse-square law. I also want to be able to show that some other factor can multiply this probability (increasing the range at which the second point can be reliably targeted).

I'm not asking anyone to do the work for me, but I described it so that someone can hopefully point me in the right direction. Can one of you resident geniuses give me some tips, or tell me what I should be studying so I can figure this out?

I don't think you are looking for a 1/r^2, but a 1/r^3 relationship.
Let R be the distance between source and target, and let r be the distance of a point from the target. Let's assume that the probability distribution F of your targetting mechanism deploying a charge at distance r from the target is radially symmetric, for example with a Gaussian profile:
[math]
F = F(r) = \alpha e^{-\frac{r^2}{2\sigma^2}}
[/math]
The normalisation factor, alpha, is just there to ensure
[math]
\int_0^\infty \int_0^{2 \pi} \int_0^\pi F r^2 \sin\theta d\phi d\theta dr = 1
[/math]
Radial symmetry is a sensible assumption. The width of the profile, \sigma, we assume to be a linear function of R. For example, you may be able to launch your charge with a given angular precision, so the lateral accuracy will be a linear function of R. Similar argument for radial precision.
The exact shape of the probability distribution isn't important, so we can make life easier by assuming a simpler profile:
[math]
F(r) = \left\lbrace \begin{array}{ll}
\alpha & \mathrm{if}\; r < \sigma\\
0 & \mathrm{otherwise}
\end{array}\right.
[/math]
i.e. a ball with constant density. Since \alpha must be chosen so that the integral over the ball sums up to 1, it should be clear that density F goes down with the cube of the ball radius, i.e.
[math]
\alpha \sim \sigma^{-3}
[/math]

Let's assume that in order to score a "hit" you have to end up within a given radius r_0 of the target. Therefore, the hit probability P is the integral of F over a ball of radius r_0. since F is constant, it follows that
[math]
P = \frac{4\pi}{3}r_0^3 F \sim \frac{4\pi}{3} \frac{r_0^3}{\sigma^3} \sim \frac{4\pi}{3} \frac{r_0^3}{R^3}
[/math]
The same is true for any other radially symmetric profile F. Your assumption of a 1/r^2 would only make sense if for example your lateral precision was a linear function of distance, but the radial targetting were always spot on, i.e. your charges would always end up on the surface of a sphere with radius R.
 

Radu094

Donator
Donator
Beta Tester
Joined
Sep 5, 2008
Messages
36
Reaction score
0
Points
0
Well, since we are in a Space forum, wouldn't it be safe to assume the said targeting would be done in orbit and outside of a gravitational field? In which case radial targeting would become irrelevant and the distribution would fall back to 1/r^2 ?
 

Spike Spiegel

New member
Joined
Feb 12, 2009
Messages
168
Reaction score
0
Points
0
Thanks Izack and Martin. I have to say Martin, you've gone quite a bit over my head there. It's as if I'm here on the ground and you're somewhere up in orbit. :) Izack to some degree as well. (I had to look up what a ~ meant, so you can imagine how my head feels after looking at Martin's equations.)

I think due to the words I chose, I may not have expressed this concept accurately enough. I'll try to be more specific. When I speak of "targeting" the second point, I'm not talking about firing a missile at it or something. This technology is a means of moving an object instantaneously between two points in space, a form of travel. You might think of this as teleporting. Since I want to impose limits on its range, and since I want the accuracy of the device to decrease as the destination's distance increases, I wanted an equation to do this in a consistent and "predictable" way. (Trying to prevent plot holes and the like.)

Given this information, does it seem that Martin's equations are not relevant to what I'm trying to do?

Either way, I'd still like to understand (as much as I can or need to) what all those fancy symbols mean. (Please excuse my ignorance.)
 

Radu094

Donator
Donator
Beta Tester
Joined
Sep 5, 2008
Messages
36
Reaction score
0
Points
0
Ok, if it is teleport rather than missile/lasers, then Martin's equations are back on the table and your targeting precission should drop by 1/r^3 as he had demonstrated
 

Notebook

Addon Developer
Addon Developer
Donator
Joined
Nov 20, 2007
Messages
11,279
Reaction score
256
Points
158

Radu094

Donator
Donator
Beta Tester
Joined
Sep 5, 2008
Messages
36
Reaction score
0
Points
0
Since I want to impose limits on its range, and since I want the accuracy of the device to decrease as the destination's distance increases, I wanted an equation to do this in a consistent and "predictable" way. (Trying to prevent plot holes and the like.)

Just as a starting point, you could start with 2 assumptions:

1.The accuracy of the device decreases with the cube of the distance to destination

StandardError(distance) = constant * distance^3

2. Pick a specific [ame="http://en.wikipedia.org/wiki/Standard_deviation"]STD [/ame]error value (3 meters) at a specific distance (50 meters). STD error will be the difference vector between your desired targeted point and your actual arrival point. This implies that things intended to be transported 50 meters will end-up in:

69% -within 47 - 53 meters (50 +/- 3)
95% -within 44 - 56 meters (50 +/- 2*3)

then you can compute your constant as :

StandardError (50 meters) = 3 meters = constant * 50^3

=> constant = 2.4e-5

You can then tweak the values untill you reach the accuracy you need for your device...
 

JEL

Addon Developer
Addon Developer
Joined
Apr 23, 2008
Messages
674
Reaction score
0
Points
0
Location
in the cold Denmark
Website
www.jelstudio.dk
Since I want to impose limits on its range, and since I want the accuracy of the device to decrease as the destination's distance increases, I wanted an equation to do this in a consistent and "predictable" way.

Maybe you could do this:

You know the maximum range of your device, let's assume it's 100 km.
X = 100

Then calculate your result this way: X minus distance = result

Your result will then be 100 when you have 100% probability, and below zero when you have zero probability.

And one last thing; this method will give you 50% probability when your distance is exactly half of the device's maximum range, so the function is linear and not exponential (in case that has a bearing in your scifi-story).

(Funny math-symbols not included. They can be purchased as extra :) )
 
Last edited:

Quick_Nick

Passed the Turing Test
Donator
Joined
Oct 20, 2007
Messages
4,073
Reaction score
156
Points
103
Location
Tucson, AZ
Just for fun, there's the scale that some teachers use as grade curve.
Square root of the linear percentage, times ten.
0 stays 0, 100 stays 100, 36 becomes 60, 81 becomes 90, etc.
 
Last edited:
Top