The correct formula is:

DeltaN = (N-R)+S

Where:

N = the number of bikes you own
R = the number of bikes you do not ride at least once every 6 months
S = the number of ridden bikes you are prepared to sell

(..and DeltaN = change in number of bikes)

So, in my case – I own 7 bikes, of which I ride all 7 at least once every 6 months. I can not bring myself to sell a single one of them.

Therefore, N does not change.

(Which sux, cos I’ve got a real hankering in the loins at the moment for a Yeti ASR-5 carbon )

it does make things difficult for people who currently own zero bikes, though?

you would sell them all for a million quid though…..

so we need a price variable…?

wwaswas… yes, correct. The model assumes that saturation has been achieved!

Maybe:

DeltaN = (N-R+I)+S

Where:

N = the number of bikes you own
R = the number of bikes you do not ride at least once every 6 months
I = The number of extra bikes you could ride at least once every six months
S = the number of ridden bikes you are prepared to sell

Is better?

Occam’s Razor. What’s wrong with N+1?

So, I have 3 bikes and 2 get ridden regularly but I’m not willing to sell the one that gets ridden rarely.

(3-1)+0 = 2

But I have 3 bikes. 😕

Iz confused…

Occam’s Razor. What’s wrong with N+1?

Only valid in theoretical situations where £=

lol at Rickos

@ Rickos

So, I have 3 bikes and 2 get ridden regularly but I’m not willing to sell the one that gets ridden rarely.

(3-1)+0 = 2

But I have 3 bikes.

Therefore, you should have 2 bikes. 😛

Conversely, if I was willing to sell the one I don’t use much (1994 Rock Lobster) the new formula would give me an answer of 3. So, do I sell it or not? This new formula sucks!

Rickos, a valid point. Just buy another bike (n+1)

I have nothing constructive to add to this thread, except to post a picture of my new Yeti ASR5c in order to prove to the OP that his reasoning is therefore incorrect and the correct number of bikes is still N+1

4-2+0 = 2

Kewl been trying to justify another bike. Now I need to buy 2 new bikes.

@ meesterbond

But. It. Just. Looks. Sooooo…… Badass. 😯 Want. One.

Approx how much for that build, if you don’t mind me asking? 😉

oooh, so the more bikes I don’t ride the more I’m allowed to buy?

Okay, given the feedback so far.. I think I may have finally cracked it:

DeltaN = (sqrt(N*R))+I+S

Where:

N = the number of bikes you own
R = the number of bikes you do not ride at least once every 6 months
I = The number of extra bikes you could ride at least once every six months
S = the number of ridden bikes you are prepared to sell

I think this one’s foolproof now…. 😀

Due to finances I’m working on the N-1 formula at the minute 🙁

(sqrt(4×2))+1+0 = 3.82
I don’t have space for 3 and a bit more bikes. Plus I’d have to buy some niche too. FS + 29er + roadbike + a shelf full of spares making up approx 0.8 of a bike?

Maybe S should be the number of bikes you are prepared to sell regardless of whether you ride them or not?

you do need to include the number of bikes your partner thinks that you own in there somewhere too?

A bit more than I’d intended to spend… it’s the ‘Race’ build from Bromley Bikes with additional Reverb, CK headset and Hope/Stans wheels.

Oooops… 😳

I now realise that ‘deltaN’ should actaully be ‘OptimalN’ – the optimal number of bikes.

Now heading to Bromley bikes website 8)

….now drooling at the thought of an ASR-5 c ‘pro’ build

…..now realising that the ‘race’ build exceeds the theoretical maximum amount I would ever consider spending on a bike ever.

…I would be terrified of ever crashing it, or letting it out of my sight for more than a few seconds.

Hmmmm..

As a middle-management type this is all a bit complicated for me, isn’t there an excel I could download somewhere to work it out?

…so, finally:

OptimalN = (sqrt(N*R))+I+S

DeltaN = OptimalN – N

So the full formula for change in bike number (DeltaN) is:

DeltaN = ((sqrt(N*R))+I+S)-N

N = the number of bikes you own
R = the number of bikes you do not ride at least once every 6 months
I = The number of extra bikes you could ride at least once every six months
S = the number of ridden bikes you are prepared to sell

So for me it is (the square root of 1*0) + 0 + 0)- 1

Edit – got my maths wrong. I am happy with my one bike, but I come out as -1??

Sorry, I think it should be this way round then! 😳

DeltaN = N-((sqrt(N*R))+I+S)

Correct… and foolproof!

I come out as:

7-((sqrt(7*0))+0+0) = 7 😀

R = the number of bikes you do not ride at least once every 6 months

Can somebody tell me how many bikes there are, so I can work this out?

But I haven’t got a road bike and I want one. What’s how many other bikes I have got to do with it? N + 1 works for me (and I’ve just sold two) (The ones I never rode).

So the more bikes I am willing to sell, the higher my optimalN? Should it not be -S?

Sums seem flawed.
If i flog the 2 I never ride and want a FS, I’m only allowed to buy 17% of a FS.
But after I’ve sold those 2 and redo the sums I’m allowed -1 bikes 😕

The thing is, I’d sell any of my bikes if the price offered was high enough. I offered a bloke my pub bike for 6grand the other day, and if he’d had cash, I’d have happily walked home.

So the formula should include a factor for avarice.