 View Single Post  3rd June 2012, 06:53 AM #2324 Robert Muehlenkamp New Blood   Join Date: May 2012 Posts: 18 Originally Posted by SnakeTongue noun the practice of taking someone elses work or ideas and passing them off as ones own. [link]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. No, genius, I refuted Mattogno's conclusions applying a) his own calculation method and b) assumptions regarding average weights that, unlike his, were realistic. Originally Posted by SnakeTongue Did you use Carlo Mattognos model and method or not? See above. Originally Posted by SnakeTongue Which is your own model and method? Mattogno's method was good enough for me, but your method may turn out more precise results - if one doesn't make the hilarious mistake you made, that is. Originally Posted by SnakeTongue I have already demonstrated in my original post: Did you refute or did you apply Carlo Mattognos model and method? See above. Originally Posted by SnakeTongue 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 kg 1.70 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. [link] Looks like a weight below 45 kg at 1.60 meters height is a strong indication of anorexia, but further factors have to be present for anorexia to be diagnosed. This means that a person weighing less than 45 kg at 1.60 meters height (i.e. more than 4 kg below what is still considered normal weight at that height according to the BMI table) may be but is not necessarily anorexic. Such person is certainly underweight, and underweight may result from anorexia (maybe it it usually does in prosperous communities where people tend to be well-fed) or from malnourishment due to insufficient food availability (which I presume is the usual cause in communities with low food intake, such as Jewish ghetto communities in Poland in the early 1940s). And I strongly doubt that every underweight person (whether due to anorexia or to malnourishment brought on by circumstances) looks as bad as the skeleton on your photograph. That must be an extreme case of anorexia, close to or below the lower limit of underweight according to the BMI table, which is 38 kg. Originally Posted by SnakeTongue (...) 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). [link][indent]In general, the men responded to semi-starvation by reducing their activity levels. They became tired, weak, listless, apathetic and complained of a lack of energy. (...) [link] Thanks for listing some factors that may have contributed to the passivity of ghetto Jews when they were made to board the trains to the extermination camps, to their high mortality on the transports and to the passivity of the survivors when they reached the camps, even though most of them knew or at least suspected that they were going to be killed. What does your source tell us about the height of the subjects and their weights prior to and after six months of semi-starvation? What it doesn't tell us, unless I missed something, is that the subjects were so weak that they couldn't move without external help, as you claimed earlier. Originally Posted by SnakeTongue The method is deceitful because induce the reader to believe the human body volume variation is absolute proportional to the body mass variation. You may call the method rudimentary, but caling it "deceitful" suggests your self-projecting paranoia. Besides, it was the method applied by your guru Mattogno. If you want to call Mattogno deceitful, be my guest. Originally Posted by SnakeTongue You applied an imaginary mass value to a hypothetical model which was only defined to estimate the capacity of a mass grave. What "imaginary mass value"? Originally Posted by SnakeTongue With the imaginary mass you obtained the hypothetical density of the mass grave. I guess you mean this: Quote: According to the Body Measurement Index table, a person with a height of 1.60 meters is underweight at 38 to 48 kg. Assuming that the average weight of adult Jews in Polish ghettos at the time was in between the upper and the lower value of what the BMI table considers underweight, it would be (38+48) ÷ 2 = 43 kg. According to Mattogno's "other tables", the weight of an adult is 2.76 times that of a child up to 14. This relation would mean a weight of 43 ÷ 2.76 = 15.6 kg for ill-fed or starving children in Polish ghettos. Rounding up the latter value, a group of two adults and one child 14 years and younger from a Jewish ghetto in Poland would thus weigh (43+43+16)/3 = 34 kg on average, instead of the 55.1 kg calculated by Mattogno. The average weight of deportees to Bełżec was probably even lower as children made up a higher proportion of deportees from Galicia, at least 42.1 % According to Mattogno's formula, 420 ÷ 34 = 12.4 (12) corpses with this average weight could fit into 1 cubic meter of grave space. --- Originally Posted by SnakeTongue Then, ignoring a scientific established method, you transformed the density into corpse per cubic meter. If I ignored a "scientific established method" (which would that be?), so did your guru Mattogno. I only followed his reasoning. Originally Posted by SnakeTongue 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. I'm not sure what the poet is trying to tell me here, but I guess he's referring to this calculation: Quote: Now to Mattognos reference weight based on "experimental data" (6 adults a 70 kg per cubic meter = 420 kg per cubic meter). Alex Bay calculated the space that would be occupied by a human being having the measurements of proportions of Leonardo Da Vinci's "Vetruvian Man", and concluded that 91,000 corpes with the proportions of the "Vetruvian Man" and an assumed height of 68 inches (1.73 meters) could have fit into 8,502 cubic meters of grave space - 10.7 (11) per cubic meter. The ideal weight of a person 1.73 meters high would be 66 kg for men and 62 kg for women. Taking the lower value, 10.7 human bodies with the measurements and weight of an ideal adult person 1.73 meters high would have a weight of 10.7 x 62 = 663.40 kg, instead of Mattogno's 420 kg. Using the former value as a reference, the unrealistically high weights assumed by Mattogno for an adult+adult+child group, i.e. (70+70+25,4) ÷ 3 = 55.13 kg, would mean 663.40 ÷ 55.13 = 12.03 (12) corpses per cubic meter. With the more realistic weights for malnourished Polish ghetto Jews mentioned above, the average would be 663.4 ÷ 34 = 19.51 (20) corpses per cubic meter. The reasoning of this is simple: if 663.40 kg of body weight corresponding to 10.7 corpses with a weight of at least 62 kg (actually I should have used 66 kg and thus obtained a higher total weight, as Bay's model was for a male body, so the 663.40 kg are conservative) fit into one cubic meter of grave space, then 663.40 kg of body weight corresponding to 19.51 people with a weight of 34 kg fit into the same amount of grave space. Originally Posted by SnakeTongue Yes, that is right. If you have a better assumption is up to you to present it. I don't know what exactly you mean by a "better assumption", but I have no problem using your calculation method instead of the more rudimentary method underlying my above-mentioned results. Without, of course, making the stupid mistake you made Originally Posted by SnakeTongue (...) post which I have already published in the JREF forum (...) [link]Bodies per cubic meter is not a measurement. 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 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. Kindly spare me and our readers the patronizing blather, and if you're now arguing that your original formula was inappopriate, please say so loud and clear instead of trying to blame me for the supposed inadequacy of your original calculations, which I merely reproduced in an Excel sheet. The following is from this thread's post 1792, written by you: Quote: The formula can be applied to the Holocaust Controversies estimations of 3 adults with a total mass (a) of 129Kg (3*43Kg), 4 children with a total mass (b) of 64Kg (4*16Kg) and 1 toddler with a total mass (c) of 16Kg (1*16Kg): {x = a/(a+b+c)*0.44/3, y = b/(a+b+c)*0.44/4, z = c/(a+b+c)*0.44/1} {x = 129/(129+64+16)*0.44/3, y = 64/(129+64+16)*0.44/4, z = 16/(129+64+16)*0.44/1} x~0.0905263, y~0.0336842, z~0.0336842 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. Your calculation method was not bad in principle, but you messed up badly in one respect, and that makes your calculation results worthless. I'll explain: I first noticed and pointed out a conspicous oddity in your calculation results: while you claimed that no more than 14 people with an average weight of 34 kilograms could be squeezed into one cubic meter, Charles Provan had managed to squeeze 8 people with an average weight of 33.25 kg into 0.44 cubic meters, which is the equivalent of 18 people per cubic meter (and there would have been room for more if test subjects had been dead and not needed to breathe). What could 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)? Things looked even more odd when I reproduced your very own calculations on an Excel spreadsheet: Item_Provan's test group_Test group with 3 adults à 43 kg and 5 children à 16 kg (a) Number of bodies in Provan's test group_8_8 (b) Volume m³ of Provan's box_0.44_0.44 (c) Volume m³ per body = (b) ÷ (a)_0.06_0.06 (d) Concentration of bodies per cubic meter_18.18_18.18 ST's formula: (e) Total weight of adults kg_174_129 (f) Total weight of children (1) kg_85_64 (g) Total weight of children (2) kg_7_16 (h) Total weight of test group kg_266_209 (i) Average weight of test person kg_33.25_26.13 (j) Volume occupied by adult m³_0.095940_0.090526 (k) Volume occupied by child (1) m³_0.035150_0.033684 (l) Volume occupied by child (2) m³_0.011579_0.033684 (m) Total volume m³ (j)+(k)+(l)_0.142669_0.157895 (n) Average volume per person = (m) ÷ 3_0.047556_0.052632 (o) Persons per m³ = 1 ÷ (n)_21.03_19.00 How could it be that the bodies of Provan's test group occupied 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 were lower than those of the former? Using your calculations for the hypothetical test group with 3 adults à 43 kg and 5 children à 16 kg, I did the following execises: 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). The results of these exercises were quite amazing, not to say hilarious: Item_43/16 scenario_21.5/8 scenario_86/32 scenario (e) Total weight of adults kg_129_65_258 (f) Total weight of children (1) kg_64_32_128 (g) Total weight of children (2) kg_16_8_32 (h) Total weight of test group kg_209_105_418 (i) Average weight of test person kg_26.13_13.06_52.25 (j) Volume occupied by adult m³_0.090526_0.090526_0.090526 (k) Volume occupied by child (1) m³_0.033684_0.033684_0.033684 (l) Volume occupied by child (2) m³_0.033684_0.033684_0.033684 (m) Total volume m³ (j)+(k)+(l)_0.157895_0.157895_0.157895 (n) Average volume per person = (m) ÷ 3_0.052632_0.052632_0.052632 (o) Persons per m³ = 1 ÷ (n)_19.00_19.00_19.00 As I confirmed by further calculations applying the same random factors to the adult and child weight values, it doesn't matter whether you double the weights, cut them in half or multiply them by or divide them through any given factor. 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, of course, means that your calculations are deeply flawed, to put it politely. Now, why is that your formula leads to such obviously mistaken results? Where did your mess up? Your mistake was that you used the same volume, 0.44 cubic meters, for Provan's test group (175+85+7 = 266 kg) and for your hypothetical test group with the weights I had assumed (3*43 + 4*16 + 16 = 209 kg), while it is obvious that the latter group, with a lower total weight, could have been placed in a box smaller than the box of Provan's test. Your calculations for the hypothetical test group should have been based on a box smaller than Provan's box in the same proportion that the weight of the hypothetical text group is smaller than that of Provan's test group: 0.44 m³ x 209÷266 = 0.345714 m³ With this lower box volume, your calculations for the 3*43 kg + 5*16 kg hypothetical test group would have turned out the following: (e) Total weight of adults kg_129.00 (f) Total weight of children (1) kg_64.00 (g) Total weight of children (2) kg_16.00 (h) Total weight of test group kg_209.00 (i) Average weight of test person kg_26.13 (j) Volume occupied by adult m³_0.071128 (k) Volume occupied by child (1) m³_0.026466 (l) Volume occupied by child (2) m³_0.026466 (m) Total volume m³ (j)+(k)+(l)_0.124060 (n) Average volume per person = (m) ÷ 3_0.041353 (o) Persons per m³ = 1 ÷ (n)_24.18 You would then have concluded that a group consisting of 2 adults à 43 kg and one child weighing 16 kg occupied the following amount of space in m³: Volume adult in m³_0.071128 Volume adult in m³_0.071128 Volume child in m³_0.026466 Total volume group in m³_0.168722 Average volume group member in m³_0.056241 Bodies per m³_17.780763 Available grave space in m³_21,310 Number of bodies that could be buried in available grave space_378,908 That would have been the correct calculation. You could also have made things easier for yourself by simply considering the following: 1. Provan's experiment gives us 266 kg in 0.44 m³, which is the equivalent of 604.545455 kg/m³. 2. 604.545455 kg correspond to 17.780749 bodies with an average weight of 34 kg. 3. Thus 17.780749 bodies with an aveage weight of 34 kg fit into 1 cubic meter of grave space. Now let's raise the weights in order demonstrate the irrelevance of your "anorexic" blather. Let's assume that adult ghetto Jews were just below normal weight threshold of the BMI table for persons 1.60 meters tall, i.e. that they weighed not 43 but 48 kg on average. According to Mattogno's "other tables", the weight of an adult is 2.76 times that of a child up to 14, so let assume that children weighed 48 ÷ 2.76 = ca. 17 kg on average. We would thus have an average weight of 38 kg for a ghetto population of which two-thirds were adults and one-third were children. Again we do the above calculation: 1. Provan's experiment gives us 266 kg in 0.44 m³, which is the equivalent of 604.545455 kg/m³. 2. 604.545455 kg correspond to 15.909091 bodies with an average weight of 38 kg. 3. Thus 15.909091 bodies with an aveage weight of 38 kg would have fit into 1 cubic meter of grave space, if the bodies of the Belzec victims had all been placed into the graves at the same time. The bodies were placed in the graves not all at the same time but over a period of ca. 8 months (March to November 1943), so decomposition must be taken into account as a factor increasing available grave space as it made the corpses in the graves lose volume. The number of "fresh" corpses per cubic meter calculated above, 15.9, is higher than the number of "fresh" corpses per cubic meter I considered in the model that led me to the following conclusion (emphasis added): Quote: Modeling the effects of corpse decomposition on the amount of grave space available at Bełżec should ideally be done on the basis of a day-by-day or at least month-by-month breakdown of the 434,508 deportees delivered at that camp according to the Höfle Report. Unfortunately no such breakdown is available. The next best thing is a table in Appendix A of Arads study on the Reinhard(t) camps that adds up to a higher number (513,142, according to my summation) and allows for a day-by-day breakdown of this number, albeit with certain assumptions and the inaccuracies inevitably resulting from such assumptions. Based on this table, I modeled a scenario of mass grave space management at Bełżec taking into account the loss of body volume due to decomposition, the results being that even 513,142 dead bodies could have been buried in 20,670 cubic meters of burial space (the volume of the burial graves according to Prof. Kolas investigation results, see section 2.1) considering decomposition-related grave space economy, and that it was therefore also possible to bury the much lower number of documented deportees to Bełżec (434,508) in the same burial space. The model assumed a density of 14.8 non-decomposed corpses per cubic meter, which means that with the density calculated above (19.51 per cubic meter) the saving of burial space due to decomposition would be even higher. And we haven't yet taken into consideration the presumable further stretching of graves by top-down partial burning of the buried corpses, which was reported by Pfannenstiel and corroborated by Cornides. Not to mention the fact that the mass graves discovered by Prof. Andrzej Kola were not necessarily the only mass graves that existed at Belzec extermination camp. Air photo analysis by Alex Bay suggests the presence of further graves. If Bay is right, that would make the above calculations rather theoretical.     