How Does Shooter Accuracy Affect Probability of Hitting a Target?

[zeeshahmad]

Homework Statement

A rifle shooter aims at a target, at a distance d, but has an accuracy controlled by a probability density:

\rho(\phi)=\frac{1}{2\Phi}

\phi\in(-\Phi,\Phi)​

where \phi is the angle achieved and is bounded by the small angle \Phi.

(Refer to attachment for diagram)

Calculate the probability density for where the bullet strikes the target, \tilde{p}(x). (I've done this.. I think)

If a target is set up with a width of 2d, with success H say, being hitting the target and failure M say, being missed the target, calculate and depict the probability of hitting,

P(H;\Phi)
as a function of \Phi for fixed d and D with D=tan(\theta)
Hint: Be careful about when \Phi=\phi

Homework Equations

p(a,b)=\int{dx p(x)}

The Attempt at a Solution

For the bit I've done,
I got \tilde{p}(x)=\frac{2\Phi}{\pi}

I don't know where to start on the next bit, could someone give me a hint please?

 

[Simon Bridge]

For best results, you should show your reasoning for the bit you've done.
How did you work that out?
What does x represent in p(x)? How is it related to $\phi$?
What is D and d (and $\theta$)?

As to your question:
You have a probability density function - how would you normally go about calculating a probability from that?

 

[zeeshahmad]

For the bit I've done:
P(a, b)=\int_a^b \! p(x) \, \mathrm{d} x.
P(x, x+\delta x)=\delta\theta\frac{1}{\pi}=p(x) \delta x (I got this from what our lecturer did in a question with a person throwing balls randomly toward a wall.)
so,
p(x)=\frac{1}{\pi} \frac{d\theta}{d x}

now, tan \theta = \frac {x}{D}

so, \frac{d\theta}{dx}= \frac {D}{D^2+x^2}

so, p(x) = \frac{1}{\pi} \frac {D}{D^2+x^2}

then, \tilde{p}(x)=p(-D tan\Phi, D tan\Phi)

= 2 \int_0^{Dtan \Phi} \! \frac{1}{\pi} \frac {D}{D^2+x^2}

= \frac{2\Phi}{\pi}

D is the distance from the shooter to the target. (shown in attached diagram)

 

[zeeshahmad]

I would integrate it over the interval I want the probability for ?

 

[Simon Bridge]

D = the distance to the target (you did that - good)
d = the radius of the target
x = radius to where a bullet went at the target distance - so that $$p(x,x+\delta x)=\frac{2\Phi\delta x}{\pi}$$... this is what you are telling me?

Check - if $\tilde{p}(x) is correct then$\int_{-X}^X \tilde{p}(x)dx=1 : X=D\tan\Phi## ... ?
(That's if I understand your thinking correctly - note: if x is a radius where did the negative values come from?)

Hint: what is the probability density if the rifle is very inaccurate indeed and $\Phi=\pi/2$ (90°)?

I think you need to get a better feel for how the statistics behave:

The image below shows a sample of 10000 shots from a flat angular distribution (top) the angle is in units of $\Phi$ and they have been grouped into lots of $\Phi/100$ so that there is an expectation of 100 shots in each histogram.

Below that is the same 100 shots according to where they hit the target. (Distance is in units of X.)
Notice how the distribution is not longer flat? Instead, you'd get bullet holes clustered closer together near the center and spreading out as you get further away.

I think it would help if you thought about what each of the equations you are writing are saying.

Note: still don't know where $\theta$ comes from... since $\phi$ is the trajectory angle, perhaps $\theta$ is the polar angle (from the x axis) to the resulting hole in the target?

 

[haruspex]

P(x, x+\delta x)=\delta\theta\frac{1}{\pi}=p(x) \delta x (I got this from what our lecturer did in a question with a person throwing balls randomly toward a wall.)

The trouble with carrying equations over from another context is that you might not have changed everytrhing that needs changing. There should be a \Phi in there somewhere, no?

now, tan \theta = \frac {x}{D}
so, \frac{d\theta}{dx}= \frac {D}{D^2+x^2}
so, p(x) = \frac{1}{\pi} \frac {D}{D^2+x^2}
then, \tilde{p}(x)=p(-D tan\Phi, D tan\Phi)
= 2 \int_0^{Dtan \Phi} \! \frac{1}{\pi} \frac {D}{D^2+x^2}
= \frac{2\Phi}{\pi}

Some confusion here. The OP provides \tilde{p}(x) as the prob density function for X. In the above, you've mostly written p(x) for that. That's fine, but then you introduce \tilde{p}(x) as something else - apparently as the integral of the pdf over its entire range, which is therefore not even a function of x.
That you got something other than 1 for that should tell you your equation P(x, x+\delta x)=\delta\theta\frac{1}{\pi} is wrong. Can you see how to correct it?

 

[zeeshahmad]

In all honesty, guys, I've just tried to collect information from here and there to solve this.
I'm not even sure if \tilde{p} or p are just different symbols or \tilde{p} has a special meaning?

I also can't distinguish the need for three angles, like Simon mentioned.

Also, I don't know what P(x, x+\delta x)=\delta\theta\frac{1}{\pi}=p(x) \delta x or p(x,x+\delta x)=\frac{2\Phi\delta x}{\pi} means.

I'm sorry for not being clear about this earlier.

 

[zeeshahmad]

That's fine, but then you introduce \tilde{p}(x) as something else - apparently as the integral of the pdf over its entire range, which is therefore not even a function of x.

\tilde{p}(x) is mentioned first in the question, but I do not know why it is mentioned instead of p(x). Perhaps a typo?

 

[haruspex]

In all honesty, guys, I've just tried to collect information from here and there to solve this.
I'm not even sure if \tilde{p} or p are just different symbols or \tilde{p} has a special meaning?

I also can't distinguish the need for three angles, like Simon mentioned.

Also, I don't know what P(x, x+\delta x)=\delta\theta\frac{1}{\pi}=p(x) \delta x or p(x,x+\delta x)=\frac{2\Phi\delta x}{\pi} means.

I'm sorry for not being clear about this earlier.

In that case, let's start from scratch so that you come to understand it. Try to answer the following in sequence:
1. Suppose on a given shot the angle error is theta. How far off target will the shot be in terms of distance (x)?
2. Same as (1), but now for an angle θ+δθ? Call this distance x+δx.
3. Obtain an expression for δx as a function of θ and δθ that's a good approximation when δθ is small.
4. What is the probability of an angle error in the range (θ, θ+δθ)?
5. What therefore is the probability of a distance error in the range (x, x+δx)?
You should then be able to write down the pdf for x.

 

[Simon Bridge]

let's start from scratch so that you come to understand it

What he said - I'll just watch for now.
I don't know why there should be three angles either - you were the one who introduced the third angle symbol ;)

 

[zeeshahmad]

I don't know why there should be three angles either - you were the one who introduced the third angle symbol ;)

Oh crap, I did!

Haruspex, you were write about copying from another context.

Ok, so

1. I think I'm looking for x=D tan(\theta)
2. x+\delta x=D tan(\theta + \delta \theta)
3.
Algebraically, \delta x=D(tan(\theta+\delta\theta)-tan(\theta))
Using expansion approximation of "tan(a+x)" - using wolfram alpha:
\delta x=D sec^{2}(x)\delta \theta

I am not exactly sure what you mean by 4. and 5.

 

[haruspex]

Oh crap, I did!

Haruspex, you were write about copying from another context.

Ok, so

1. I think I'm looking for x=D tan(\theta)
2. x+\delta x=D tan(\theta + \delta \theta)
3.
Algebraically, \delta x=D(tan(\theta+\delta\theta)-tan(\theta))
Using expansion approximation of "tan(a+x)" - using wolfram alpha:
\delta x=D sec^{2}(x)\delta \theta

I am not exactly sure what you mean by 4. and 5.

Good so far. For (4), you are told the probability density function for the error angle, namely, that it is uniform over a certain range. If the range (θ, θ+δθ) is within that range, what is the probability that the error angle lies within it?

 

[zeeshahmad]

that would be
\frac{\theta+\delta \theta}{\Phi} ?

 

[haruspex]

that would be
\frac{\theta+\delta \theta}{\Phi} ?

No, that would be the prob of the angle error being in the range (-θ-δθ, θ+δθ).
Try integrating the pdf over the range (θ, θ+δθ).

 

[zeeshahmad]

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi

=\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi

=\frac{1}{2} \left[ ln(\phi) \right] (i can't put the limits here with latex)

=\frac{1}{2} (ln(\theta + \delta \theta)-ln(\theta))

= ln(\sqrt{1+\frac{\delta \theta}{\theta}})

 

[Simon Bridge]

In latex markup you can formal special functions like the natural logarithm and add the limits like this:

=\frac{1}{2} \ln|\phi| \bigg |_\theta^{\theta+\delta\theta}​

to give you (put above inside doubled "$" signs) $$=\frac{1}{2} \ln|\phi| \bigg |_\theta^{\theta+\delta\theta}$$

But ... surely $\Phi$ is a constant in the integration? If so: how do you get the log?
Did you mix up $\phi$ and $\Phi$?

 

[haruspex]

zeeshahmad, I've come to suspect there are several typos in the OP which are confusing things. I would guess:
- the distance to the target is D, not d.
- θ is defined by d = D tan (θ)
- the hint should be to take care when $\Phi$ < θ.
A further confusion is that you freely interchanged θ and $\phi$, and I followed suit.
To clarify:

$\Phi$ is a constant angle
$\phi$ is a variable angle of error, which is anywhere in the range ($-\Phi$, $\Phi$)
D is the distance to the target
x is a variable distance error, x = D tan($\phi$)
θ is the limit of angle error for a 'hit'
d is the limit of distance error for a 'hit', d = D tan(θ)​

Is that all correct?

 

[zeeshahmad]

All of that is correct, just that the last two statements are not in the question (not that I disagree with them), and \theta is only mentioned (1) in the hint and (2) in d = D tan(θ)

But the hint says exactly as mentioned in OP.

 

[zeeshahmad]

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi

=\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi

=\frac{1}{2} \left[ ln(\phi) \right] (i can't put the limits here with latex)

=\frac{1}{2} (ln(\theta + \delta \theta)-ln(\theta))

= ln(\sqrt{1+\frac{\delta \theta}{\theta}})

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi

=\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi

=\frac{d\delta}{2\Phi} ...

 

[haruspex]

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi

=\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi

=\frac{d\delta}{2\Phi} ...

Almost right. Where you wrote $d \delta$ at the end you meant $\delta \theta$, right?
So to my step 5: what is the probability that the distance error lies in the range (x, x+δx) where, you may remember, $x = D tan(\phi)$ and $x+\delta x = D tan(\phi+\delta \phi)$?

 

[zeeshahmad]

Yes.

$$P(x, x+\delta x)=P(Dtan(\phi), Dtan(\phi+\delta\phi))$$

$$\int_x^{x+\delta x} \!{p(x)} = \int_{Dtan(\phi)}^{Dtan(\phi+\delta\phi)} <-not-sure-what-goes-in-there$$

 

[haruspex]

Step 5 is the easiest!
But first, to avoid confusion in future, I must once again clean up the use of theta versus phi. Step 4 should have referred to phi, not theta, so the answer to it becomes $\frac{\delta \phi }{ 2\Phi}$
The distance error is in the range $(x, x+\delta x)$ iff the angle error is in the range $(\phi, \phi+\delta \phi)$, right? And step 4 computed the second of those, so the answer to step 5 is?

 

[zeeshahmad]

the same expression

 

[haruspex]

the same expression

Precisely. But to get the pdf, you need to rewrite it in terms of x and δx.

 

[zeeshahmad]

$$\frac{1}{2\Phi} (arctan(\frac{x+\delta x}{D})-arctan(\frac{x}{D}))$$

 

[haruspex]

$$\frac{1}{2\Phi} (arctan(\frac{x+\delta x}{D})-arctan(\frac{x}{D}))$$

Yes, but you can simplify that for small deltas. What's the derivative of arctan?

 

[zeeshahmad]

for $$arctan(\frac{x}{D})$$

it is $$\frac{D}{D^2+x^2}$$

but why do we need the derivative?
also, for small delta x, the expression is 0, isn't it?

(Edit:sorry for messing up the variables again, I typed small d's before)

 

[zeeshahmad]

oh I get it,

$$\int{p(x) dx} = \frac{1}{2\Phi} (arctan(\frac{x+\delta x}{D})-arctan(\frac{x}{D}))$$

if that is right, we would then differentiate it..
but what about the \delta x ? How do we take care of that?

 

[haruspex]

for $$arctan(\frac{x}{D})$$

it is $$\frac{D}{D^2+x^2}$$

but why do we need the derivative?
also, for small delta x, the expression is 0, isn't it?

Basic process of differentiation: f(x+δx) ≈ f(x)+f'(x)δx
Put $f(x) = arctan(\frac{x}{D})$ and see what happens.

 

[zeeshahmad]

i get

$$p(x)=\frac{1}{2\Phi}(\frac{D}{D^2+x^2})$$

 

[haruspex]

Looks right to me.
So now, what about P(H;Φ)? Hint: there's an easier way than using p(x).

 

[zeeshahmad]

is it 1 when, theta is greater than Phi?

 

[Simon Bridge]

is it 1 when, theta is greater than Phi?

... when the target area is bigger than the spread of the bullets? Well... yes. That would be part of the answer.

You are thinking along the right lines though...
How about when $\theta < \Phi$?
How would you determine the probability that $\phi$ is inside a certain angle?