The Formula For The Optimum Number Of Basses...

[lowregisterhead]

An accurate formula has finally been developed for calculating the optimum number of basses you can own:

It is: x-1, where 'x' is the number of basses which when reached, will cause your wife/husband/partner to leave you.


(With thanks to Robert Elms, whose version of the formula was for calculating the number of bikes you could own.)

[paul_5]

Duly noted. I'll look into transposing this formula to fit a pedalboard model...

[BigRedX]

Rubbish. As any fule kno the optimum number of basses to own is n+1 where n = the number of basses you already have.

[jonno1981]

I know someone who had a firearms license (he used to keep shotguns) and he would keep all his music gear locked up in the secure cupboard he had. His excuse was that legally he was the only person who was allowed entry into the room! He had loads of stuff! He would also only buy guitars in certain colours and rotate the 1 or 2 he had in the house. Genius! His wife never looked at them closely enough to realise he had different ones out every other week as they were all the same colour!

Edited June 22, 2013 by jonno1981

[stef030]

[i]the formula is [/i]
[i]to have as many as you like and if the old kitchen viper[/i]
[i]says anything you just get rid of her, [/i]
[i]buy another bass,then get another girl [/i]


[i]problem solved[/i]

cheers
stef



ps
I am of to brighton to look for another jazz lol

[TimR]

So it is x+1<n-1 for positive integers. Where n is a constant fixed variable dependant on environmental emotional conditions and x is a variable that can increase and decrease freely dependant on volume and pressure of the GAS.

[4 Strings]

[quote name='TimR' timestamp='1371923447' post='2119704']
So it is x+1<n-1 for positive integers. Where n is a constant fixed variable dependant on environmental emotional conditions and x is a variable that can increase and decrease freely dependant on volume and pressure of the GAS.
[/quote]

But this doesn't account for rate of change.

If the increase in the number of basses per unit of time exceeds 'N' (Notice factor) then the game's up.

For safe operation:

n/t<N

where
n=no. of basses
t = time (year, for example, in extreme cases, month)

If, by joining Basschat, there is an increase in the rate at which the number of basses increases then we have an n[sup]2[/sup] situation and there is no hope for bass player or partner as n[sup]2[/sup]/t will quickly be larger than almost any reasonable value of N

[Lowender]

I've spent many years working on this formula. I finally got it down to owning all the basses that sound like no other and none of the basses that are just "variations" of those sounds.

The list, for me, is as such...

P
J
MM
Rick
Fretless
Ibanez
Hofner
Obligatory 5 string
upright
And the Squier Deluxe active J (Yeah, I know -- an odd choice, but a wonderfully unique instrument).

Edited June 23, 2013 by Lowender

[TimR]

[quote name='4 Strings' timestamp='1371987005' post='2120301']


But this doesn't account for rate of change.

If the increase in the number of basses per unit of time exceeds 'N' (Notice factor) then the game's up.

For safe operation:

n/t<N

where
n=no. of basses
t = time (year, for example, in extreme cases, month)

If, by joining Basschat, there is an increase in the rate at which the number of basses increases then we have an n[sup]2[/sup] situation and there is no hope for bass player or partner as n[sup]2[/sup]/t will quickly be larger than almost any reasonable value of N
[/quote]

Good point so:

dx/dt=x - n[sup]2[/sup]/t + 2 for positive integer values.

Edited June 23, 2013 by TimR

[4 Strings]

[quote name='TimR' timestamp='1372012395' post='2120662']
Good point so:

dx/dt=x - n[sup]2[/sup]/t + 2 for positive integer values.
[/quote]

See, now I failed A Level maths, so I was thinking more n<mt+c

where n = number of basses
m = rate at which you can accumulate without notice
t = time
c = the number of basses with which you started when you met her.

This would conclude that if the rate at which you accumulate without notice becomes dangerously high you can start again with a new 'er indoors with a higher number of basses with which you started, confirming the previous post which suggested this.

[TimR]

I think we're confusing n[sup]2[/sup]/t with m.

Probably need to redefine terms and relationships from the start.

x is number of basses you currently have.
n is the number of basses you can have before you lose your house.
t is time
m is the maximum rate of change.

n = mt + x
m < n[sup]2[/sup]/t
n - 1 < x + 1 or n < x + 2

we need to find the maximum value of n and the time t that it occurs in terms of m.

It's 9 o'clock on Sunday night and I need more wine to solve that.

Edited June 23, 2013 by TimR

[4 Strings]

[quote name='TimR' timestamp='1372017930' post='2120753']
It's 9 o'clock on Sunday night and I need more wine to solve that.
[/quote]

I think that's best!

[Dad3353]

Please remember, students, that the original question was 'Optimal', not 'Maximum'. I know these are often the same, or similar, but for full mathematical accuracy it would be a shame to confuse the two. I would suggest that the term we need to find be 'O', and that the time scale be limited to 1L, where L = one lifetime. I realise that the solution may verge on the infinite, but I'm hoping there's a 'real' solution without needing to invoke Chaos theory. Of course when 'O' exceeds 'M', then Chaos is the constant which results.
Good luck, and tidy up the bottles afterwards, please.

Edited June 23, 2013 by Dad3353

[icastle]

[quote name='lowregisterhead' timestamp='1371888003' post='2119092']
It is: x-1, where 'x' is the number of basses which when reached, will cause your wife/husband/partner to leave you.
[/quote]

There's also an interesting variant: x+s, where 's' is directly related to the speed at which you want the person to leave and is proportional to bass count.

[lowregisterhead]

Wonderful work, gentlemen. Last night, in a moment of rare clarity, I presented our findings to She Who Must Be Obeyed, and within minutes was handed a sheet of paper with the following typed on it:

"Bollocks."

I'm not certain what that means, but I'm fairly sure it's not an equation.

[4 Strings]

[quote name='lowregisterhead' timestamp='1372059770' post='2120995']
Wonderful work, gentlemen. Last night, in a moment of rare clarity, I presented our findings to She Who Must Be Obeyed, and within minutes was handed a sheet of paper with the following typed on it:

"Bollocks."

I'm not certain what that means, but I'm fairly sure it's not an equation.
[/quote]

It's the answer to the question "What would you say if someone commented that you, as a wife, limited the number of basses in the house?"

[TimR]

[quote name='Dad3353' timestamp='1372026107' post='2120873']
Please remember, students, that the original question was 'Optimal', not 'Maximum'. I know these are often the same, or similar, but for full mathematical accuracy it would be a shame to confuse the two. I would suggest that the term we need to find be 'O', and that the time scale be limited to 1L, where L = one lifetime. I realise that the solution may verge on the infinite, but I'm hoping there's a 'real' solution without needing to invoke Chaos theory. Of course when 'O' exceeds 'M', then Chaos is the constant which results.
Good luck, and tidy up the bottles afterwards, please.
[/quote]

An interesting postulation. I think for the purposes of this thread we can consider optimal as being just below the point that undesirable results occur. By defining n in terms of m we can safely assume that in some cases m will be very small and other cases (lonley single men with loads of money) infinite.

Of course some, like iCastle may use the results to their advantage by finding n and reach it before time t and thus exceed the boundary limits of the equation.

[Phil Adams]

Surely this is simple.
X +1
Where X = the number of basses you can hide from your wife-girlfriend-lover-boyfriend-partner-thing.

[Dave_the_bass]

I've been giving this some serious thought and believe that many very important variables have been missed from this:

O = optimum number of basses
n = number of basses owned when co-habitation commenced
m = time in months of co-habitation
a = average cost of desired basses
F = amount of disposable income that can be diverted in to a separate account per month without being noticed
L = number of levers in the mortice lock on the music room door.

So, my carefully considered equation

O = n+((mF)/a) L)+2

Essentially this means that if the partner has permanent access to the music room then you can only ever have number of basses that is two higher than when you moved in together.

worked example

n = 2
m = 100
a = 1500
F = 200
L = 0

O = 2 + ((100x200)/1500)x 0) + 2
O = 2 + ((20000)/1500) x 0) + 2
O = 2 + ((13.333) x 0) + 2
O = 2 + (0) + 2
O = 4

The addition of a 5 lever mortice lock to the music room door would dramatically increase the optimum number of basses.

That is all!

[4-string-thing]

[quote name='stef030' timestamp='1371909513' post='2119450']
[i]the formula is [/i]
[i]to have as many as you like and if the old kitchen viper[/i]
[i]says anything you just get rid of her, [/i]
[i]buy another bass,then get another girl [/i]


[i]problem solved[/i]

cheers
stef



ps
I am of to brighton to look for another jazz lol
[/quote]
Yup, women come and go, but a bass is for life!

[TimR]

[quote name='Dave_the_bass' timestamp='1372075225' post='2121190']
I've been giving this some serious thought and believe that many very important variables have been missed from this:

O = optimum number of basses
n = number of basses owned when co-habitation commenced
m = time in months of co-habitation
a = average cost of desired basses
F = amount of disposable income that can be diverted in to a separate account per month without being noticed
L = number of levers in the mortice lock on the music room door.
...

That is all!
[/quote]

you're confusing this by using non standard variables and not the standards as defined above.

your variables should be as follows:

n = optimum number of basses
x = number of basses owned when co-habitation commenced
M = time in months of co-habitation
a = average cost of desired basses
F = amount of disposable income that can be diverted in to a separate account per month without being noticed
L = number of levers in the mortice lock on the music room door.

Thus the part of your equation "((mF)/a) L)" would become ((MF)/a)L) where ((MF)/a)L) = m. So substituting into the original equation we now have:

x is number of basses you currently have.
n is the number of basses you can have before you lose your house.
t is time
m is the maximum rate of change.

n = mt + x
m < n[sup]2[/sup]/t
n - 1 < x + 1 or n < x + 2
m = (MF)/a)L

[Myke]

So after reading this.. I have come to the conclusion.. that I won't be getting a wife..

[TimR]

[quote name='Myke' timestamp='1372159268' post='2122253']
So after reading this.. I have come to the conclusion.. that I won't be getting a wife..
[/quote]

I don't even want to start thinking about creating a formula with no limits.

[Dave_the_bass]

Assuming no use of credit facilities; the formula with no limits is

NtF /a

Where:

Nt is the number of months you've been playing bass

F and a are as before

[MiltyG565]

X is non-existant for me... HURRAY!