Thread: Why HHT is important or is not?

Originally Posted by proximus26

Correct,
this what exactly happen to hair when i touched blade. however after one shave I got HHT 4. I guess pure luck and end of story. I never got more than HHT5 since then.

In order for the HHT6 to succeed you must be an 91st level blademaster and sing the last two stanzas of the "honemeister's chant" backwards.

am so beside myself,got a 3.7687 today,moving on up to the big time

Originally Posted by pixelfixed

am so beside myself,got a 3.7687 today,moving on up to the big time

You're doing better than I am. I can't find rope thin enough to tie a noose small enough to hang my hair with.

Originally Posted by jaswarb

HHT 6 has only been report twice. It only occurs when the razor is on it's way and the aforementioned hair cuts itself down.

Yes, and don't forget, it must be Leap Year for that to occur!

Hang on now, is this root in or root out? Because as we all know, root in (RI) HHT3 is equivalent to root out (RO) HHT1, but RO HHT 2 only approaches RI HHT5 at asymptotic nebulosity.

James.

Edit: I like that phrase - I am going to make it my new user title.

Originally Posted by Jimbo

Hang on now, is this root in or root out? Because as we all know, root in (RI) HHT3 is equivalent to root out (RO) HHT1, but RO HHT 2 only approaches RI HHT5 at asymptotic nebulosity.

James.

Edit: I like that phrase - I am going to make it my new user title.

You know, sometimes I like jumping straight to the end of a thread just to see if I can make any kind of sense out of what folks are talking about.

I fear this may have been a tactical error on my part this time...

Originally Posted by jaswarb

In order for the HHT6 to succeed you must be an 91st level blademaster and sing the last two stanzas of the "honemeister's chant" backwards.

I've been questing a lot, but those damn monsters just keep failing to drop the mythical HHT6 skill. This is world of warcraft all over again.
Though I have to say that my hairline keeps receding the more I shave. Does that count?

Originally Posted by Jimbo

Hang on now, is this root in or root out? Because as we all know, root in (RI) HHT3 is equivalent to root out (RO) HHT1, but RO HHT 2 only approaches RI HHT5 at asymptotic nebulosity.

James.

Edit: I like that phrase - I am going to make it my new user title.

Don't forget the exact angle of the hair in relation to the edge in a 3D space.
And the length of the hair that protrudes beyond the edge.
As soon as we include those factors, we have to agree on whom to shave completely bald so that we all have a supply of equivalent hairs.

It occurrs to me that the HHT is the 'Fahrenheit' scale of shaving: Based on 2 reference points, one of which is random, and one of which is wrong.

Well, might I then suggest a nomenclature for HHT exactitude?

HHT-x = subjective ordinal-scaled rating, x = 0, 1, 2, ...
Root In = RI
Root Out = RO
Ratio of hair length protruding beyond edge = r
Length of hair = L
Weight profile of hair as a function of length from root = w(l)
Wind Speed (knots) = Wi
Heart Rate of Tester = HR
Angle of approach (horizontal) = \theta
Angle of approach (vertical) = \phi

Now with these variables we may formulate what I would call a load of Basic Unadjusted Local Lebesgue-Space Holistically Integrated Tests as follows:

\int_{0}^{rL} [ (icos(theta) + sin(phi)) * (1/(\sqrt{2\pi}) * \exp{-Wi/2 * (HR^2 (w(l) - 42)/HHT-x}^\delta(RI) ] dW,

where W follows a Wiener process and of course the integrand is non-anticipating.

James.

Originally Posted by Jimbo

Well, might I then suggest a nomenclature for HHT exactitude?

HHT-x = subjective ordinal-scaled rating, x = 0, 1, 2, ...
Root In = RI
Root Out = RO
Ratio of hair length protruding beyond edge = r
Length of hair = L
Weight profile of hair as a function of length from root = w(l)
Wind Speed (knots) = Wi
Heart Rate of Tester = HR
Angle of approach (horizontal) = \theta
Angle of approach (vertical) = \phi

Now with these variables we may formulate what I would call a load of Basic Unadjusted Local Lebesgue-Space Holistically Integrated Tests as follows:

\int_{0}^{rL} [ (icos(theta) + sin(phi)) * (1/(\sqrt{2\pi}) * \exp{-Wi/2 * (HR^2 (w(l) - 42)/HHT-x}^\delta(RI) ] dW,

where W follows a Wiener process and of course the integrand is non-anticipating.

James.

Well said! Those of us with some science background really appreciate just how subjective the HHT can be unless ALL the variables are understood.



Nevertheless, until we come up with something better as a pre-shave assessment of sharpness, then it will have to do.