Explain the Circular Error Probable formula mentioned in this article
Metin Bektas in this blog post writes about how we could calculate the probability (p) of hitting a target by a missile with a given accuracy (measured as CEP) as below:
p = 1 – exp( -0.41 · R² / CEP² )
Definition of CEP (Circular Error Probable) taken from the blog post:
An important quantity when comparing missiles is the CEP (Circular Error Probable). It is defined as the radius of the circle in which 50 % of the fired missiles land.
How was the formula for p derived?
probability statistics
asked Sep 13 '16 at 5:36
ardsrk
1013
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
Metin Bektas in this blog post writes about how we could calculate the probability (p) of hitting a target by a missile with a given accuracy (measured as CEP) as below:
p = 1 – exp( -0.41 · R² / CEP² )
Definition of CEP (Circular Error Probable) taken from the blog post:
An important quantity when comparing missiles is the CEP (Circular Error Probable). It is defined as the radius of the circle in which 50 % of the fired missiles land.
How was the formula for p derived?
probability statistics
asked Sep 13 '16 at 5:36
ardsrk
1013
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06
add a comment |
Metin Bektas in this blog post writes about how we could calculate the probability (p) of hitting a target by a missile with a given accuracy (measured as CEP) as below:
p = 1 – exp( -0.41 · R² / CEP² )
Definition of CEP (Circular Error Probable) taken from the blog post:
An important quantity when comparing missiles is the CEP (Circular Error Probable). It is defined as the radius of the circle in which 50 % of the fired missiles land.
How was the formula for p derived?
probability statistics
asked Sep 13 '16 at 5:36
ardsrk
1013
Metin Bektas in this blog post writes about how we could calculate the probability (p) of hitting a target by a missile with a given accuracy (measured as CEP) as below:
p = 1 – exp( -0.41 · R² / CEP² )
Definition of CEP (Circular Error Probable) taken from the blog post:
An important quantity when comparing missiles is the CEP (Circular Error Probable). It is defined as the radius of the circle in which 50 % of the fired missiles land.
How was the formula for p derived?
probability statistics
probability statistics
asked Sep 13 '16 at 5:36
ardsrk
1013
asked Sep 13 '16 at 5:36
ardsrk
1013
asked Sep 13 '16 at 5:36
ardsrk
1013
asked Sep 13 '16 at 5:36
ardsrk
1013
asked Sep 13 '16 at 5:36
ardsrk
1013
1013
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06
add a comment |
Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06
Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06
add a comment |
1 Answer
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The formula is a variation on the cumulative density function (CDF) of the Rayleigh probability distribution as found here:
$$F(x,sigma) = 1 - exp(-frac{x^2}{2sigma^2})$$
where $sigma$ is the mode of the distribution. To represent the formula in terms of the radius corresponding to a 50% probability, or $CEP^2$, you would need to use the formula as given by user "Did" above. I agree that the referenced blog post is incorrect.
answered Jan 5 '17 at 20:36
user404485
1
add a comment |
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1 Answer
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The formula is a variation on the cumulative density function (CDF) of the Rayleigh probability distribution as found here:
$$F(x,sigma) = 1 - exp(-frac{x^2}{2sigma^2})$$
where $sigma$ is the mode of the distribution. To represent the formula in terms of the radius corresponding to a 50% probability, or $CEP^2$, you would need to use the formula as given by user "Did" above. I agree that the referenced blog post is incorrect.
answered Jan 5 '17 at 20:36
user404485
1
add a comment |
The formula is a variation on the cumulative density function (CDF) of the Rayleigh probability distribution as found here:
$$F(x,sigma) = 1 - exp(-frac{x^2}{2sigma^2})$$
where $sigma$ is the mode of the distribution. To represent the formula in terms of the radius corresponding to a 50% probability, or $CEP^2$, you would need to use the formula as given by user "Did" above. I agree that the referenced blog post is incorrect.
answered Jan 5 '17 at 20:36
user404485
1
add a comment |
The formula is a variation on the cumulative density function (CDF) of the Rayleigh probability distribution as found here:
$$F(x,sigma) = 1 - exp(-frac{x^2}{2sigma^2})$$
where $sigma$ is the mode of the distribution. To represent the formula in terms of the radius corresponding to a 50% probability, or $CEP^2$, you would need to use the formula as given by user "Did" above. I agree that the referenced blog post is incorrect.
answered Jan 5 '17 at 20:36
user404485
1
The formula is a variation on the cumulative density function (CDF) of the Rayleigh probability distribution as found here:
$$F(x,sigma) = 1 - exp(-frac{x^2}{2sigma^2})$$
where $sigma$ is the mode of the distribution. To represent the formula in terms of the radius corresponding to a 50% probability, or $CEP^2$, you would need to use the formula as given by user "Did" above. I agree that the referenced blog post is incorrect.
answered Jan 5 '17 at 20:36
user404485
1
answered Jan 5 '17 at 20:36
user404485
1
answered Jan 5 '17 at 20:36
user404485
1
answered Jan 5 '17 at 20:36
user404485
1
1
add a comment |
add a comment |
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Using 2D (isotropic) gaussian distributions.
– Did
Sep 13 '16 at 5:49
@Did could you please elaborate. What is the constant -0.41?
– ardsrk
Sep 13 '16 at 14:34
This is odd, the formula in the blog post gives 33% chances to hit the disk of radius CEP. To get 50% chances, one should use $p=1-exp(-0.69cdot R^2/text{CEP}^2)$ instead. Typo?
– Did
Sep 13 '16 at 16:03
The value $0.69$ ($=ln2$) is confirmed by the French WP page.
– Did
Sep 13 '16 at 16:06