What happens if . . . (Full Version)

All Forums >> [Community Discussions] >> Ask a Mistress



Message

PeonForHer -> What happens if . . . (2/16/2010 5:52:15 AM)

A Domme, whose buzz out of being dominant is about the effect she's having on a sub, gets together with a sub whose buzz is entirely about what she's getting out of dominating him?   Has it ever happened?

It struck me recently that, though 'do-me-ism' amongst subs is an oft-mentioned annoyance for Dommes, perhaps some subs go too far the other way.  You're watching him for the effect you're having on him doing x, y or z to him - but he doesn't care about x, y or z - he's only watching you to see what effect your doing x, y or z is having on you.

Just a thought.  Quite likely another example of my over-thinking and under-doing, too.  Re that last, please use this thread to ramble all over the place, if you like.
;-)



onlyfreelycaged -> RE: What happens if . . . (2/16/2010 6:40:00 AM)

I've had this problem from both sides of the slash.... as a top it annoys the hell out of me. I expect pepple I'm playing with to have likes, dislikes, and express them... other wise it's no fun doing stuff he hates..

now.. as for being a sub.. yes, I have this problem with my *favorate* Dom... when he gets board and asked me what I'd like.. and I answer "what ever you want to do with me.."



VaguelyCurious -> RE: What happens if . . . (2/16/2010 7:09:14 AM)

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.



dreamerdreaming -> RE: What happens if . . . (2/16/2010 7:47:51 AM)

Not with me, no. Things are made more enjoyable when others enjoy them with you of course, but I don't start things with patrners that don't already have D/s needs and desires that coincide nicely with mine.

Things that don't work for both of us, don't work for either of us. Those ideas fall by the wayside fast. The minute I find out he's not into a certain activity, for whatever reason, I'm completely turned off by the idea as well. We're in this for mutual pleasure and fulfillment.

The idea that he wouldn't enjoy a certain activity, makes it very unnattractive to me. [:'(]

There are ideas that fell away quickly in the beginning, that we may revisit later. Because yes, it is yummy to think that someone would do (insert activity here) just because they get off on your reaction. But for right now, we're really turned off to the idea of doing things that we aren't equally as enamored with. Because there are so many things we both adore doing, and so little time.



GloriousMorning -> RE: What happens if . . . (2/16/2010 11:08:41 AM)

I'm sure it happens, but I can't say I've ever experienced that myself.

I enjoy what I do with my partner because for me, it is a journey of discovery of mutual enjoyment and pleasure in an exchange of power and energy, to create mutually enjoyable experiences and exploration of our selves and one another.

I don't think I would enjoy being with a partner who was to focused on meeting my needs, or who only sought satisfaction by proxy of my pleasure. Even as a Dominant, I enjoy giving pleasure as much as being on the receiving end. In fact most times I find enjoyment in discovering exactly what it is my partner enjoys, and discovering ways to control it, not take away, but to control his pleasure buttons. Other times, I just want him to hit my pleasure button. ;)

The dynamics I usually have with my partner is more of a give/take, connector/receiver type of energy exchange. I have times when I am very focused on the reactions I am creating in my partner,  and others when he is very focused on mine. I would be more likely to focus on what we both find pleasure in, that brings out equal amounts of enjoyment, so we both feed into eachother's responses in a continuous loop of sorts. I suppose that's why they call it an energy or power exchange.



LadyAngelika -> RE: What happens if . . . (2/16/2010 4:50:12 PM)

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.



Hot! ;-)

I like boys who get off on my reaction. It is a good match for me. That said, I do get off on their reactions too, and unless they are made of stone, I will find a way to get a reaction out of them. Trust me. ;-)

- LA



PeonForHer -> RE: What happens if . . . (2/16/2010 5:20:13 PM)

Pfft.  Sometimes it doesn't take much, LA.  I met a domme in a pub once.  She told me to go and buy her a packet of cigs from the newsagent down the road.  Put it this way: it was a painful trot there and back.



velt -> RE: What happens if . . . (2/16/2010 5:20:21 PM)

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.



PeonForHer -> RE: What happens if . . . (2/16/2010 5:22:57 PM)

You know, you could have just said "You're thinking about it too much, Peon"

That was absolutely hilarious, though. [:D]



LadyAngelika -> RE: What happens if . . . (2/16/2010 5:29:07 PM)

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.



Brilliant!! Who is our new found friend without a profile? Me likey!!

- LA



shallowdeep -> RE: What happens if . . . (2/16/2010 9:34:37 PM)

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...

[image]local://upfiles/324704/6A91D608093F4441903DD82917DAF35B.jpg[/image]



VaguelyCurious -> RE: What happens if . . . (2/17/2010 1:15:45 AM)

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...

[image]local://upfiles/324704/6A91D608093F4441903DD82917DAF35B.jpg[/image]


Dear Shallowdeep,

I don't know who you are or where you are from but I have just fallen in love with you.

Yours,

Vague



VaguelyCurious -> RE: What happens if . . . (2/17/2010 1:21:22 AM)

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?



subinchico -> RE: What happens if . . . (2/17/2010 3:08:53 AM)

The results of a positive feedback loop between Ss and Sd would color up a sceen like a non-stop bleeder and increased heart rate.  Ever heard of that sceen played out? 



MsStarlett -> RE: What happens if . . . (2/17/2010 4:27:10 AM)

Velt and Shallow, you both are just TO much fun!

But yes, Peon, it does happen.  It just causes a bad match.  The concept of the "Power Exchange" (for me) is based on that feed back.  The 'power' or 'electricity' or 'excitement' has to flow back and forth.  That's one reason I personally am not overly into blindfolds, gags or hoods... I want to SEE and HEAR the reaction and experience it.  Although my dear Wall-e wants to be gagged for our next session, I bet he can still make some wonderful whimpering noises.



PeonForHer -> RE: What happens if . . . (2/17/2010 4:45:02 AM)

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?



Actually, you're both wrong.  You have both failed to factor in the speed of light, squared.



LadyAngelika -> RE: What happens if . . . (2/17/2010 5:05:01 AM)

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...

[image]local://upfiles/324704/6A91D608093F4441903DD82917DAF35B.jpg[/image]


Engineers do it in diagrams ;-) Agreeing with VC, I love it!

- LA



LadyAngelika -> RE: What happens if . . . (2/17/2010 5:11:44 AM)

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?



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



velt -> RE: What happens if . . . (2/17/2010 9:19:54 AM)

Es is the expressivity of the sub, and Pd the perceptivity of the domme. Since Sd is the stimulus provided by the domme to the sub, aren't Sd(n) and Ss(n) in the wrong places in the diagram?

(On my part, I made a few typos and it should've said "That for any (Es x Pd) x (Ed x Ps) < 1 that Ss(x) and Sd(x) approrach 0 as x increases,")

Regarding BIBO, making sure the output would be unbounded was half the fun, of course.



velt -> RE: What happens if . . . (2/17/2010 9:46:07 AM)

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



Regarding additional separate instances of stimulation, I suppose I was thinking along the lines that additional actions of the domme would be contained within the expressivity (Ed) of the domme in the following cycle. In PeonForHer's original post, the sub "doesn't care about x, y or z - he's only watching you to see what effect your doing x, y or z is having on you.", so instances of physical stimulation in this case would be about as exciting for the sub as stubbing their toe (not at all... for me anyway). This does bring up the question of what form the initial stimulation (Si) took to begin with, though. Assuming additional sources of outside stimulation (Si) could be added, do you think they would be better modeled by incorporation into the existing loop or expressed as the summation of all discrete instances of initial stimulation for a given point in time?



Page: [1] 2   next >   >>

Valid CSS!




Collarchat.com © 2025
Terms of Service Privacy Policy Spam Policy
0.046875