What a shame that Carlo Mattogno didn’t have the assistance of your mathematical genius, because my "bizarre" model and "worst" method are based on the method applied by Mattogno. From the blog "Mattogno, Graf & Kues on the Aktion Reinhard(t) Mass Graves (3)":
A bit hysterical, are we? If
I refute Mattogno using his own calculation method, that's not plagiarism where I come from. Maybe you should look up the term.
Nick Terry said:
Until you get it through your thick skull that Roberto Muehlenkamp used multiple methods to arrive at his estimates, then there is really nothing to discuss here.
Nick was presumably referring to my establishing grave capacity first without taking into consideration the effect of decomposition and then taking into account that effect.
Calling me names already in your first message? Quite the hysteric, really.
Previously you admitted your model and method was based on Carlo Mattogno’s model and method and now you are admitting you just refuted Carlo Mattogno’s model and method.
Did you use Carlo Mattogno’s model and method or not?
Which is your own model and method?
Bravo, genius. Now explain the components of your so-much-better formula, tell us what assumptions your figures are based on, and let's see the results of your so-much-better calculation.
(...)
Sounds nice, but what's the genius trying to tell us here? Instead of throwing theoretical wisdom around, show us your calculations of the volume of space occupied by three dead people weighing 43, 43 and 16 kg. (...)
I have already demonstrated in my original post:
Snaketongue said:
Using the Holocaust Controversies distribution of 2 adults and 1 child the total volume of all bodies is:
V = 0.0905263 + 0.0905263 + 0.0336842
V = 0.2147368m^3
The average body volume of 2 adults and 1 child is 0.07158 cubic meters. Thus a 21,310 cubic meters burial pit would hold up to 297,713 bodies of adults and children with an average weight of 34 kilograms.
But first tell us that "Revisionist" guru Mattogno's method (which I also applied) is the worst possible method. I'd like to read that from you. I shall then send Mattogno an e-mail with your opinion about his method.
Did you refute or did you apply Carlo Mattogno’s model and method?
(...) I don't think such a person would look like the skeleton in your picture, which seems to be emaciated rather than just underweight. I'd say that person's weight is below the lowest range of Untergewicht according to the BMI table. But you are free to demonstrate the contrary.
How to tell if you are anorexic:
Are you fat?
Was your last period more than three months ago?
Does your weight fall below these sort of numbers:
metric:
1.60 metres: 45 kg1.70 metres: 51 kg
1.80 metres: 57 kg
English:
5 foot: 90 lbs
5 foot 6: 108 lbs
6 foot: 129 lbs
If you answered “yes” to at least two of these questions then there is a good chance you are anorexic. See your doctor to be sure.
http://abagond.wordpress.com/2008/01/22/anorexia/
I'd say that depends on the definition of "underweight" and on the degree of undernourishment. A person 1.60 meters tall weighing 48 kg (just 1 kg below what would still be normal weight) would hardly require "constant assistance to move". If you think you can demonstrate that a person this tall weighing 5 kg less would require such assistance, fire away. Throwing unsubstantiated generalities around doesn't cut it.
(...) As the 6 months of semistarvation progressed, the volunteers exhibited many physical changes, including gastrointestinal discomfort; decreased need for sleep; dizziness; headaches; hypersensitivity to noise and light; reduced strength; poor motor control; edema (an excess of fluid causing swelling); hair loss; decreased tolerance for cold temperatures (cold hands and feet); visual disturbances (i.e., inability to focus, eye aches, "spots" in the visual fields); auditory disturbances (i.e., ringing noise in the ears); and paresthesias (i.e., abnormal tingling or prickling sensations, especially in the hands or feet).
http://www.joyproject.org/overcoming/starvation.html
Now I'm getting curious. How exactly am I supposed to have "misinterpreted" Charles Provan's experiment (whose results I merely compared with the results I had arrived at by a different method, which you call the "worst" method), and how exactly is my method supposed to be "deceitful"? Please be more specific. And bear in mind that what you are accusing me of regarding my method you are also accusing your guru Mattogno of, for my method is no different from his.
(...) Am I supposed to have been "deceitful" to the disadvantage of my own argument?
The method is deceitful because induce the reader to believe the human body volume variation is absolute proportional to the body mass variation. You applied an imaginary mass value to a hypothetical model which was only defined to estimate the capacity of a mass grave. With the imaginary mass you obtained the hypothetical density of the mass grave. Then, ignoring a scientific established method, you transformed the density into “corpse per cubic meter”. Thus you propose that density variation could estimate the volume of your lower mass odd model. Since you applied unknown values to the Charles A Bay hypothesis, you extrapolated the results.
Still waiting for your superior decomposition model, my friend. And please show us where you got the idea that every grave was "closed and sealed approxmately each 8 days". I hope for you that you didn't just divide the number of days corresponding to eight months by the number of graves discovered by Prof. Andrzej Kola.
Yes, that is right. If you have a better assumption is up to you to present it.
I'd love to have a chat with you on the CODOH forum, but unfortunately the CODOH moderator, Mr. Jonnie "Hannover" Hargis, is so afraid of me that he banned me for good from that lovely place (after deleting many of my posts, last time I was there).
(...)
Another thing: If you're also posting your wisdom on CODOH, kindly link the CODOH thread to the discussion here. Thanks.
(...) How do you explain so large a difference in concentration (14 vs. 18 per cubic meters) despite a low difference in average weight (34 vs. 33.25 kg)?
Bodies per cubic meter is not a measurement.
(...) bodies of Provan's test group occupy less volume on average than those of the hypothetical test group with 3 adults à 43 kg and 5 children à 16 kg, even though the total and average weights of the latter test group are lower than those of the former:
(...)
(g) Total weight of children (2) kg_
7_
16
(...)
(m) Total volume m³ of (j)+(k)+(l)_0,142669_0,157895
Average volume per person = (m) ÷ 3_0,047556_0,052632
How do you explain this?
On item
(g) you regarded a toddler as a child. This explains why the first average volume is higher than the second average volume on item
(m).
If my formula is applied with an appropriate low mass for the toddler (5Kg), the total average volume is:
V = x+y+z
{x = 129/(129+64+5)*0.44/3, y = 64/(129+64+5)*0.44/4, z = 5/(129+64+5)*0.44/1}
x~0.0955556, y~0.0355556, z~0.0111111
V = 0.0955556+0.0355556+0.0111111
V = 0.1422223m^3
x = 0.1422223m^3 /3
x =
0.0474074m^3
(...)
Now let's look only at the 3*43kg + 5*16 scenario and do the following exercises:
a) cut the weights in half (21.5 kg for adults, 8 kg for children)
b) double the weights (86 kg for adults, 32 kg for children).
(...)
(h) Total weight of test group kg_209_105_418
(...)
(m) Total volume m³ (j)+(k)+(l)_0.157895_0.157895_0.157895
(...) As long as you use the same multiplier or divider for both and the relation between adult weight and child weight is not changed, the average volume occupied by one person in the test group will always be 0.052632 m³, and the concentration of bodies per cubic meter will always be 19.
This is not the appropriate way to estimate capacity of a hypothetical space. The average volume of each body does not represent an accurate factor to estimate the number of bodies which a mass grave could hold. Since bodies of adults, children and toddlers have drastic difference in their absolute volume, the average volume of each group must be applied in accordance with a distribution ratio of adult per child per toddler. Instead to obtain the average volume of x, y and z and then estimate the capacity of a given space, I would use x, y and z in accordance with a body distribution ratio no lower than 2:1:1 or no higher than 3:4:1.
Item (m) of your calculation is a body distribution ratio of 1:1:1. Therefore, if you use the same multiplier or divider with my formula, the expected result is a common factor. If there is no variation in the proportional volume occupied by each group, there is no change of the average volume of each body.