CobberKane
Banned
- 706
- Apr 4, 2012
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The reports I have read don't say a 50/50 chance of knocking down a B-17 with 20 rounds - merely that the estimate is 20 20mm rounds to knock out a B17, and that 4-5 20 mm rounds from the front would down a B-17.
That's 800 rounds fired for a non-frontal assault, or 180 rounds for a frontal assault.
I'd also guess the low chance to hit comes from pilots firing out of range or at least effective range, you more rookie pilots or those that are afraid to close.
More skilled pilots would get more rounds on target, less skilled of course less.
I think I remember reading where Saburo Sakai stated they changed to head on attacks against the B-17's.
I believe the report found that it took on average 20 x 20mm hits to down a B-17. That means, of a population of B-17s that each took that degree of punishment, 50% would go down and 50% would not. Or, as applied to an individual aircraft hit by that number of shells, there was a 50/50 chance of it going down.
assigning the low hit rate to the effect of rookie pilots doesn't work,
to be fair allied fighter with gun on wing instead of nose and the lack of cannon make them really bad again bomber
Dy definition of the word "average", that is not correct. Average means one plane may go down with 10 hits, another with 20, and another with 30. The "average" here would be 20.
If you have a 50/50 chance of bringing it down with 20 rounds, the average then would not be 20, it would be higher.
.The German report being discussed (at least, the one I have) says 20 hits from 20mm shells are what is required to be certain of bringing down a four-engine bomber
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Aha! Could be a classic case of the dangers of relying on second hand information, although if the report says 20 hits are 'certain' to bring down a bomber that is obviously not so - they may be applying the term to a given high probability, like 95%. German translator anywhere?
I'm sorry Gary, but this is incorrect. The average is the value required to reach the 50th percentile, so if a hundred bombers are each hit twenty times, fifty of them would go down and fifty would not, and when single bombers were attacked they would require, on average, twenty hits to be destroyed.
Order all the values in the data set from smallest to largest.
Multiply k percent by the total number of values, n.
This number is called the index.
If the index obtained in Step 2 is not a whole number, round it up to the nearest whole number and go to Step 4a. If the index obtained in Step 2 is a whole number, go to Step 4b.
4a.Count the values in your data set from left to right (from the smallest to the largest value) until you reach the number indicated by Step 3.
The corresponding value in your data set is the kth percentile.
4b.Count the values in your data set from left to right until you reach the number indicated by Step 2.
The kth percentile is the average of that corresponding value in your data set and the value that directly follows it.
PHP:I'm sorry Gary, but this is incorrect. The average is the value required to reach the 50th percentile, so if a hundred bombers are each hit twenty times, fifty of them would go down and fifty would not, and when single bombers were attacked they would require, on average, twenty hits to be destroyed.
O.K, answer me this - 5 B-17's are shot down. The take 12,15,20,21, and 26 rounds to down. What is the average # of hits to kock down a B-17?
We are looking for the Mean here, not the Median or Mode.
If you want to do percentile standing:
Here we are getting the median really, or 20 in my example above.
There is not really a stat that is equal to the average number of hits to put it down 50% of the time that I am aware
Actually, thinking about it, your stat reminds me of a Yogi Berra quote: "Baseball is 90% mental, the other half is physical."
Great stuff, mhuxt. If 20 x 20mm shells were ''certain'' (in fact, highly likely) to knock down a B-17, then the average would be much lower, perhaps around ten. It would be great to know how the figure was arrived at, but that might be too much to ask (sigh).
oops - double post
Not so simple, Typhoons with 4 Hispanos had no problems with LW bombers same true to Beaus and Mossies with four Hispanos in nose.
Juha
i think they mean it will take average of 20 x 20mm shells that hit to bring down a bomber
btw there 3 kind of average : median , mean , and mode , it more likely that in the example they using either mean or mode
actually all these 3 are average , but of different kind to deal with different sort of statisticAccording to the translation 20x20mm are certain to bring down a B-17 - not on average. And no, median and mode are not types of average, regardless of how the term might be used in common vernacular; the mean is the average, nothing else.
I'll leave the subject of applied mathematics alone now. There is nothing I've mentioned that couldn't be verified by a simple google search, in the unlikely event anyone is interested
they have different advantages and disadvantage:The mean, median and mode are types of average.