quote:

ORIGINAL: VaguelyCurious

I can think of at least one experience where the opposite happened.

I met a guy from CM (still have yet to meet any women from here, but this site is working wonders for my homoflexibility!) whose 'buzz' was mainly about my enjoyment, and it totally amplified the whole experience.

I'd do something and he'd flinch which would make me smile and then he'd catch that and moan and then I'd be laughing like a loon...

One massive feedback loop. Happy times.

quote:

ORIGINAL: velt

Clearly these circumstances have the possibility of introducing an anomaly into the sub-domme continuum.

Using the following variables:
Es - expressivity of sub
Ed - expressivity of domme
Ps- perceptivity of sub
Pd - perceptivity of domme
Si - strength of initial stimulus
Ss - strength of stimulus from sub to domme
Sd- strength of stimulus from domme to sub

Then for the sub, Ss = Si x Es x Pd , and Sd = Si x Ed x Ps for the domme, initially.
For the following round of reactions, Ss2 = Sd x Es x Pd =  (Si x Ed x Ps) x Es x Pd and Sd2 = Ss x Ed x Ps = (Si x Es x Pd) x Ed x Ps
From this it follows that Ss2/Sd = Sd2/Ss for simultaneously occurring reactions since Si x Ed x Ps x Es x Pd = Si x Es x Pd x Ed x Ps
And that Ss(x+1)/S(x)=Sd(x+1)/Ss(x) absent of additional stimuli and when Es, Ed, Ps and Pd are constant.
The implication of which is that for any (Es x Pd) x (Ed x Ps) > 1 that Ss(x) and Sd(x) approach infinity as x increases, and:
That for any (Es x Pd) x (Ed x Ps) < that Ss(x) and Sd(x) approrach 0 as x increases, and:
That for (Es x Pd) x (Ed x Ps) = 1 that (Ss(x) + Sd(x))/2 will remain at Si.
And that the initial Si has no impact on the final result.
Also, when Es + Ps and Ed + Pd are fixed, the best results will come from when Es is close to Ps and Ed is close to Pd.

quote:

ORIGINAL: shallowdeep

That really needs a diagram, I think. And that is definitely not a BIBO stable system. Still, poorly engineered positive feedback loops can be fun to watch in action...

quote:

ORIGINAL: velt

Clearly these circumstances have the possibility of introducing an anomaly into the sub-domme continuum.

Using the following variables:
Es - expressivity of sub
Ed - expressivity of domme
Ps- perceptivity of sub
Pd - perceptivity of domme
Si - strength of initial stimulus
Ss - strength of stimulus from sub to domme
Sd- strength of stimulus from domme to sub

Then for the sub, Ss = Si x Es x Pd , and Sd = Si x Ed x Ps for the domme, initially.
For the following round of reactions, Ss2 = Sd x Es x Pd =  (Si x Ed x Ps) x Es x Pd and Sd2 = Ss x Ed x Ps = (Si x Es x Pd) x Ed x Ps
From this it follows that Ss2/Sd = Sd2/Ss for simultaneously occurring reactions since Si x Ed x Ps x Es x Pd = Si x Es x Pd x Ed x Ps
And that Ss(x+1)/S(x)=Sd(x+1)/Ss(x) absent of additional stimuli and when Es, Ed, Ps and Pd are constant.
The implication of which is that for any (Es x Pd) x (Ed x Ps) > 1 that Ss(x) and Sd(x) approach infinity as x increases, and:
That for any (Es x Pd) x (Ed x Ps) < that Ss(x) and Sd(x) approrach 0 as x increases, and:
That for (Es x Pd) x (Ed x Ps) = 1 that (Ss(x) + Sd(x))/2 will remain at Si.
And that the initial Si has no impact on the final result.
Also, when Es + Ps and Ed + Pd are fixed, the best results will come from when Es is close to Ps and Ed is close to Pd.

quote:

ORIGINAL: VaguelyCurious
It's gonna converge sooner than your equations say it will (glorious as they are)-you don't hit them once and then just stop, after all, do you?

Aaaand incidentally, Si is going to increase with every stroke-although that won't affect the end value, it should affect the rate of convergence, right?

quote:

ORIGINAL: shallowdeep

That really needs a diagram, I think. And that is definitely not a BIBO stable system. Still, poorly engineered positive feedback loops can be fun to watch in action...

quote:

ORIGINAL: VaguelyCurious

quote:

ORIGINAL: velt

Clearly these circumstances have the possibility of introducing an anomaly into the sub-domme continuum.

Using the following variables:
Es - expressivity of sub
Ed - expressivity of domme
Ps- perceptivity of sub
Pd - perceptivity of domme
Si - strength of initial stimulus
Ss - strength of stimulus from sub to domme
Sd- strength of stimulus from domme to sub

Then for the sub, Ss = Si x Es x Pd , and Sd = Si x Ed x Ps for the domme, initially.
For the following round of reactions, Ss2 = Sd x Es x Pd =  (Si x Ed x Ps) x Es x Pd and Sd2 = Ss x Ed x Ps = (Si x Es x Pd) x Ed x Ps
From this it follows that Ss2/Sd = Sd2/Ss for simultaneously occurring reactions since Si x Ed x Ps x Es x Pd = Si x Es x Pd x Ed x Ps
And that Ss(x+1)/S(x)=Sd(x+1)/Ss(x) absent of additional stimuli and when Es, Ed, Ps and Pd are constant.
The implication of which is that for any (Es x Pd) x (Ed x Ps) > 1 that Ss(x) and Sd(x) approach infinity as x increases, and:
That for any (Es x Pd) x (Ed x Ps) < that Ss(x) and Sd(x) approrach 0 as x increases, and:
That for (Es x Pd) x (Ed x Ps) = 1 that (Ss(x) + Sd(x))/2 will remain at Si.
And that the initial Si has no impact on the final result.
Also, when Es + Ps and Ed + Pd are fixed, the best results will come from when Es is close to Ps and Ed is close to Pd.

It's gonna converge sooner than your equations say it will (glorious as they are)-you don't hit them once and then just stop, after all, do you?

Aaaand incidentally, Si is going to increase with every stroke-although that won't affect the end value, it should affect the rate of convergence, right?

quote:

ORIGINAL: LadyAngelika

quote:

ORIGINAL: VaguelyCurious

It's gonna converge sooner than your equations say it will (glorious as they are)-you don't hit them once and then just stop, after all, do you?

Aaaand incidentally, Si is going to increase with every stroke-although that won't affect the end value, it should affect the rate of convergence, right?

I agree with you that with Si increasing with every stroke, it will affect the rate of convergence but I would say that the end value is affected as overtime, the s builds up resistence to Si. The notion of adaptation is missing from this equation, no?

- LA