Octane
17-01-2006, 04:44 PM
Hi all,
I am not sure of this, but I don't believe there exists a Universal formula (which applies to Australia, anyway) for working out bad weather (for observing and imaging) after the ordering or purchasing of new equipment, and how it affects everyone else in the observing community.
In this post I propose two formulas.
Typically, there are two factors involved when making astronomical purchases. Firstly, there is cost. This hobby is cost-prohibitive for a lot of people. Secondly, there is size. Some equipment is big, some equipment is small. I propose that a user utilises the appropriate formula based on whether they purchase a telescope, a mount, etc. (cost) or an eyepiece, focal reducer, etc. (size).
1. COST-BASED FORMULA
Research shows that the more expensive the equipment we purchase, the more bad weather we encounter, either immediately following the order of the product, or the purchase of the product. One only has to read the iceinspace.com.au forums to find disgruntled observers.
W = sqrt(C * (E_u / E_p))
Legend:
W = weather (the higher the W, the worse the weather; the interpretation of this figure is approximately equal to hours)
C = cost of new equipment
E_u = enthusiasm to use
E_p = enthusiasm to pay
Note: E_u and E_p are ratings between 1 and 10; 1 being small and 10 being big.
Example:
Joe Bloggs purchases an 8" Schmidt-Cassegrain telescope valued at $2,895. He is extremely enthusiastic to use his new toy, and is fairly happy to part with his cash.
W = sqrt(C * (E_u / E_p))
W = sqrt(2895 * (10 / 8))
W = sqrt(2895 * 1.25)
W = sqrt(3618.75)
W = 60.156047077579823305056688781065
W = 60.16 (2 decimal places)
It therefore follows that Joe Bloggs, and the majority of other observers, consequently, will be unable to use their equipment for at least two and-a-half days after he receives his new toy.
2. SIZE-BASED FORMULA
Statistics reveal that the duration of bad weather is inversely proportional to the size of the new equipment ordered or purchased. This has been reinforced through the sheer number of observers having purchased Telrad's and eyepieces, and then not being able to test or use them.
W = ((1 / S) * (E_u / E_p)) * 100
Legend:
W = weather (the higher the W, the worse the weather; the interpretation of this figure is approximately equal to hours)
S = size of new equipment
E_u = enthusiasm to use
E_p = enthusiasm to pay
Note: S, E_u and E_p are ratings between 1 and 10; 1 being small and 10 being big.
Example:
Joe Bloggs purchases a Telrad valued at $89. He is excited to use his new toy, and is quite happy to part with his cash.
W = ((1 / S) * (E_u / E_p)) * 100
W = ((1 / 3) * (8 / 8)) * 100
W = (0.33333333333333333333333333333333 * 1) * 100
W = 33.333333333333333333333333333333
W = 33.33 (2 decimal places)
It therefore follows that Joe Bloggs, and the majority of other observers, consequently, will be unable to use their equipment for at least one and-a-quarter days after he receives his new toy.
Quod erat demonstrandum,
Humayun
I am not sure of this, but I don't believe there exists a Universal formula (which applies to Australia, anyway) for working out bad weather (for observing and imaging) after the ordering or purchasing of new equipment, and how it affects everyone else in the observing community.
In this post I propose two formulas.
Typically, there are two factors involved when making astronomical purchases. Firstly, there is cost. This hobby is cost-prohibitive for a lot of people. Secondly, there is size. Some equipment is big, some equipment is small. I propose that a user utilises the appropriate formula based on whether they purchase a telescope, a mount, etc. (cost) or an eyepiece, focal reducer, etc. (size).
1. COST-BASED FORMULA
Research shows that the more expensive the equipment we purchase, the more bad weather we encounter, either immediately following the order of the product, or the purchase of the product. One only has to read the iceinspace.com.au forums to find disgruntled observers.
W = sqrt(C * (E_u / E_p))
Legend:
W = weather (the higher the W, the worse the weather; the interpretation of this figure is approximately equal to hours)
C = cost of new equipment
E_u = enthusiasm to use
E_p = enthusiasm to pay
Note: E_u and E_p are ratings between 1 and 10; 1 being small and 10 being big.
Example:
Joe Bloggs purchases an 8" Schmidt-Cassegrain telescope valued at $2,895. He is extremely enthusiastic to use his new toy, and is fairly happy to part with his cash.
W = sqrt(C * (E_u / E_p))
W = sqrt(2895 * (10 / 8))
W = sqrt(2895 * 1.25)
W = sqrt(3618.75)
W = 60.156047077579823305056688781065
W = 60.16 (2 decimal places)
It therefore follows that Joe Bloggs, and the majority of other observers, consequently, will be unable to use their equipment for at least two and-a-half days after he receives his new toy.
2. SIZE-BASED FORMULA
Statistics reveal that the duration of bad weather is inversely proportional to the size of the new equipment ordered or purchased. This has been reinforced through the sheer number of observers having purchased Telrad's and eyepieces, and then not being able to test or use them.
W = ((1 / S) * (E_u / E_p)) * 100
Legend:
W = weather (the higher the W, the worse the weather; the interpretation of this figure is approximately equal to hours)
S = size of new equipment
E_u = enthusiasm to use
E_p = enthusiasm to pay
Note: S, E_u and E_p are ratings between 1 and 10; 1 being small and 10 being big.
Example:
Joe Bloggs purchases a Telrad valued at $89. He is excited to use his new toy, and is quite happy to part with his cash.
W = ((1 / S) * (E_u / E_p)) * 100
W = ((1 / 3) * (8 / 8)) * 100
W = (0.33333333333333333333333333333333 * 1) * 100
W = 33.333333333333333333333333333333
W = 33.33 (2 decimal places)
It therefore follows that Joe Bloggs, and the majority of other observers, consequently, will be unable to use their equipment for at least one and-a-quarter days after he receives his new toy.
Quod erat demonstrandum,
Humayun