SourMash

August 3, 2010, 11:11 PM

Can someone please explain what standard deviation is in simple terms?

SourMash

August 3, 2010, 11:11 PM

Can someone please explain what standard deviation is in simple terms?

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Zak Smith

August 3, 2010, 11:20 PM

http://en.wikipedia.org/wiki/Standard_deviation

Steve Marshall

August 3, 2010, 11:37 PM

Standard Deviation (SD) in simplistic terms is how much variation of any type of measure you can expect to get. This would be driven by a fact and modified by the population. e.g. say you have 50 30-06 cases and you measure the length of 10. You would have an Extreme Spread (ES) of say, .020". By applying statistics to this with the values of each one of your 10 piece sample, SD would now predict where the average of the rest of the 40 cases would fall. Small sample sizes can give you relevant answers but the larger the sample, the closer the results will be to the prediction. So in effect, by measuring something and applying statiscs to it, you can fairly accurately predict where the rest of your population will be.

Vacek

August 4, 2010, 12:39 AM

In short, take your data...say the velocity of 20 rounds. Average the 20 velocities. Now determine the delta of each separate velocity relative to the average. Take an average of the deltas and you basically have your standard deviation. If the data is unbiase i.e. normally distributed approximately 66% of the individual velocities' deltas fall into the standard deviation. 95% of the individual velocities' deltas would fall in 2X the standard deviation and 99% the 3X.

DickM

August 4, 2010, 09:14 AM

In short, take your data...say the velocity of 20 rounds. Average the 20 velocities. Now determine the delta of each separate velocity relative to the average. Take an average of the deltas and you basically have your standard deviation.

No, that would be the mean (or average) deviation, and if you calculate it exactly as you describe it would always be equal to zero, and therefore incorrect. The mean deviation is the average of the absolute value (i.e., change the sign of the negative deviations before summing) of the deviations.

To calculate standard deviation, the individual deltas from the mean are first squared, then summed to produce a statistic known as the variance. The square root of the variance is the standard deviation. The difference between the mean deviation and the standard deviation is that the larger deltas have more effect on the standard deviation due to the squaring.

A couple of minor related points - although the procedure described will get you to the same value, statisticians do not actually calculate standard deviation that way (not that anyone cares now that we have calculators/computers that do all the work, but I remember calculating lots of standard deviations by hand). Also, that formula will generate the standard deviation of your sample, but it's a biased estimator of the true population standard deviation, which is calculated by dividing the sums of the squared deviations by one less than the number of observations (i.e., by n-1, rather than n).

But to answer SourMash's original question, standard deviation is one of several ways to describe the variability of a number of measurements, such as shell length or velocity of a particular load. The smaller the standard deviation, the more similar the measurements are. As Vacek described, in normal populations - those that follow the classic bell-shaped curve - the standard deviation can be used to calculate the percentage of the measurements that will fall within a certain range, and (more importantly) can also be used to calculate the probability that an unknown measurement will be greater or less than a specified value. And it has many other valuable properties that are beyond this discussion.

FWIW, I've tested a number of data sets of velocities from individual loads and they have all been normal, or nearly so.

No, that would be the mean (or average) deviation, and if you calculate it exactly as you describe it would always be equal to zero, and therefore incorrect. The mean deviation is the average of the absolute value (i.e., change the sign of the negative deviations before summing) of the deviations.

To calculate standard deviation, the individual deltas from the mean are first squared, then summed to produce a statistic known as the variance. The square root of the variance is the standard deviation. The difference between the mean deviation and the standard deviation is that the larger deltas have more effect on the standard deviation due to the squaring.

A couple of minor related points - although the procedure described will get you to the same value, statisticians do not actually calculate standard deviation that way (not that anyone cares now that we have calculators/computers that do all the work, but I remember calculating lots of standard deviations by hand). Also, that formula will generate the standard deviation of your sample, but it's a biased estimator of the true population standard deviation, which is calculated by dividing the sums of the squared deviations by one less than the number of observations (i.e., by n-1, rather than n).

But to answer SourMash's original question, standard deviation is one of several ways to describe the variability of a number of measurements, such as shell length or velocity of a particular load. The smaller the standard deviation, the more similar the measurements are. As Vacek described, in normal populations - those that follow the classic bell-shaped curve - the standard deviation can be used to calculate the percentage of the measurements that will fall within a certain range, and (more importantly) can also be used to calculate the probability that an unknown measurement will be greater or less than a specified value. And it has many other valuable properties that are beyond this discussion.

FWIW, I've tested a number of data sets of velocities from individual loads and they have all been normal, or nearly so.

DickM

August 4, 2010, 10:34 AM

Sorry, but I can't resist adding one more thing about using standard deviations.

It's not enough to simply look at the standard deviation (or mean deviation, or range, or whatever statistic of dispersion you like) to determine whether one group of measurements is more variable than another. You also need to consider the magnitude of the measurements.

As an example, consider the two types of measurements I mentioned in the post above: shell length and velocity. Shell length is typically, in the US anyway, measured in inches, and velocity (it's speed, actually, but that's another discussion) is measured in feet per second. So, if you have a group of measurements of shell length they might vary from, say, 2.4 in to 2.6 in and lets say for the sake of argument that they have a mean (average) of 2.5 and a standard deviation (sd) of 0.1 in. Compare that to a group of velocities ranging from 3000 fps to 3200 fps with a mean of 3100 and an sd of 100. The sd of the velocities is much larger, but that doesn't necessarily mean they're more variable. We'd expect a group of measurements with magnitudes of 2-3 to have a smaller sd than a group of 3000 to 3200, right? Statisticians refer to that property of data as heteroscedasticity, which means that the sd is correlated (varies with) the mean - the bigger the mean, the bigger the sd is expected to be.

One way to compare apples to apples is to express the sd as a percentage of the mean, a statistic known as the coefficient of variation (CV). The CV of the shell length measurements is .1/2.5*100 = 4%, while the CV of the velocities is 100/3100*100 = ~3.2%. So, in fact the velocities are less variable even though their sd is one thousand times larger than the sd of the shell lengths.

All that may be obvious in this example, but if you're asking the question of which of your rifle/load combinations has the most consistent velocities, and one of them has velocities in the 2000 fps range while the other is in the 3000 fps range, it may be not-so-obvious that you can't simply pick the one with the lowest sd.

It's not enough to simply look at the standard deviation (or mean deviation, or range, or whatever statistic of dispersion you like) to determine whether one group of measurements is more variable than another. You also need to consider the magnitude of the measurements.

As an example, consider the two types of measurements I mentioned in the post above: shell length and velocity. Shell length is typically, in the US anyway, measured in inches, and velocity (it's speed, actually, but that's another discussion) is measured in feet per second. So, if you have a group of measurements of shell length they might vary from, say, 2.4 in to 2.6 in and lets say for the sake of argument that they have a mean (average) of 2.5 and a standard deviation (sd) of 0.1 in. Compare that to a group of velocities ranging from 3000 fps to 3200 fps with a mean of 3100 and an sd of 100. The sd of the velocities is much larger, but that doesn't necessarily mean they're more variable. We'd expect a group of measurements with magnitudes of 2-3 to have a smaller sd than a group of 3000 to 3200, right? Statisticians refer to that property of data as heteroscedasticity, which means that the sd is correlated (varies with) the mean - the bigger the mean, the bigger the sd is expected to be.

One way to compare apples to apples is to express the sd as a percentage of the mean, a statistic known as the coefficient of variation (CV). The CV of the shell length measurements is .1/2.5*100 = 4%, while the CV of the velocities is 100/3100*100 = ~3.2%. So, in fact the velocities are less variable even though their sd is one thousand times larger than the sd of the shell lengths.

All that may be obvious in this example, but if you're asking the question of which of your rifle/load combinations has the most consistent velocities, and one of them has velocities in the 2000 fps range while the other is in the 3000 fps range, it may be not-so-obvious that you can't simply pick the one with the lowest sd.

Bovice

August 4, 2010, 11:02 AM

lol

at one time I knew precisely how to calculate a standard deviation, correlation, covariance, etc.

but I've never once had to use it in the real world.

at one time I knew precisely how to calculate a standard deviation, correlation, covariance, etc.

but I've never once had to use it in the real world.

THe Dove

August 4, 2010, 11:45 AM

Torque - - can someone explain torque to me!!!! BWAHAHAHA Just kidding folks..

The Dove

The Dove

Smokey Joe

August 4, 2010, 12:02 PM

Higher math always gives me a headache.

Sour Mash--All you (and I) need to know about standard deviation is:The smaller the standard deviation, the more similar the measurements are.So, for velocities from a chronometer, for example, a small SD (say, in the teens) shows that there is very little variance between one round and the next, and you probably have a nice accurate load there. But I let the chrono's calculator do all the mathematical heavy lifting.

Someone is going to say that SD isn't very high for higher math. Well, any higher, and I get altitude sickness. :D Someone else is going to come along with a crack about a standard deviate, too! :D :D

Sour Mash--All you (and I) need to know about standard deviation is:The smaller the standard deviation, the more similar the measurements are.So, for velocities from a chronometer, for example, a small SD (say, in the teens) shows that there is very little variance between one round and the next, and you probably have a nice accurate load there. But I let the chrono's calculator do all the mathematical heavy lifting.

Someone is going to say that SD isn't very high for higher math. Well, any higher, and I get altitude sickness. :D Someone else is going to come along with a crack about a standard deviate, too! :D :D

SourMash

August 4, 2010, 09:14 PM

Thanks gentlemen, I've just started reloading and have seen this written down in some ballistic charts and I didn't understand what it meant. I haven't got that technical with my loads yet as I'm just learning, but hey, now I know!

dmazur

August 4, 2010, 11:01 PM

Torque - - can someone explain torque to me!!!!

Torque can be explained very quickly. In fact, it only takes a moment...

( :) )

Torque can be explained very quickly. In fact, it only takes a moment...

( :) )

MEHavey

August 4, 2010, 11:15 PM

http://i35.tinypic.com/214a9mh.jpg

Eb1

August 5, 2010, 12:04 AM

Torque - - can someone explain torque to me!!!! BWAHAHAHA Just kidding folks..

The Dove

To add some fun to the education. Torque is what you feel in the seat of your pants when you get on the throttle of my Harley.

Just having fun.. Thanks for the more educated posts.

The Dove

To add some fun to the education. Torque is what you feel in the seat of your pants when you get on the throttle of my Harley.

Just having fun.. Thanks for the more educated posts.

MCMXI

August 5, 2010, 12:38 AM

I find it odd that ALL of my velocities for EVERY load I've ever tested fall within the mean +/- 2 standard deviations. This is the case for MEHavey's example shown above as well. Consider this real-world .308 Win data for example. All velocities for three different loads fall with the mean +/- 2*SD.

http://128.171.62.162/hawthorn-engineering/thr/rem700_308win/range_targets/2010/02_14_10/chronograph_data.jpg

:)

http://128.171.62.162/hawthorn-engineering/thr/rem700_308win/range_targets/2010/02_14_10/chronograph_data.jpg

:)

MCMXI

August 5, 2010, 12:53 AM

Same thing for my .45-70 Govt. velocity results. 100% of the velocities fall within mean +/- 2*SD. :confused:

http://128.171.62.162/hawthorn-engineering/thr/marlin/1895/range_targets/02-20-10/chronograph_data_1.jpg

:)

http://128.171.62.162/hawthorn-engineering/thr/marlin/1895/range_targets/02-20-10/chronograph_data_1.jpg

:)

DickM

August 5, 2010, 09:16 AM

I find it odd that ALL of my velocities for EVERY load I've ever tested fall within the mean +/- 2 standard deviations.

It's likely just an artifact of your sample size not being large enough. Remember that there's only a 5% probability that any given shot will fall outside the mean +/- 2 sd range, so with only 10 shots your probability of having one outside that range is 1-(.95^10), or about 40%. With a 13-shot group you've got about a 50-50 chance of getting one beyond 2 sd from the mean. Note also, as MEHavey pointed out, that your sd is being calculated with some error due to the sample size, though those errors should be both positive and negative.

The statistics don't lie (though some statisticians do), and the properties of the normal distribution are very well understood.

ETA: I checked some of your data, and your chronograph does calculate standard deviation correctly. It's just a bit difficult to tell because of the rounding, but it appears to be providing you with the population standard deviation (which is what you want) rather than the sample standard deviation (see Post #5). For example, for your .45-70 24" barrel data, the population sd is 8.17 and the sample sd is 8.61. I also tested that particular data set for normality (Shapiro-Wilk W test, if anyone cares), and it appears to be normally distributed.

It's likely just an artifact of your sample size not being large enough. Remember that there's only a 5% probability that any given shot will fall outside the mean +/- 2 sd range, so with only 10 shots your probability of having one outside that range is 1-(.95^10), or about 40%. With a 13-shot group you've got about a 50-50 chance of getting one beyond 2 sd from the mean. Note also, as MEHavey pointed out, that your sd is being calculated with some error due to the sample size, though those errors should be both positive and negative.

The statistics don't lie (though some statisticians do), and the properties of the normal distribution are very well understood.

ETA: I checked some of your data, and your chronograph does calculate standard deviation correctly. It's just a bit difficult to tell because of the rounding, but it appears to be providing you with the population standard deviation (which is what you want) rather than the sample standard deviation (see Post #5). For example, for your .45-70 24" barrel data, the population sd is 8.17 and the sample sd is 8.61. I also tested that particular data set for normality (Shapiro-Wilk W test, if anyone cares), and it appears to be normally distributed.

MCMXI

August 5, 2010, 04:22 PM

It's likely just an artifact of your sample size not being large enough.

OK ... then take all 30 shots fired in the .308 Win data set above. Here are the results shown below. Bare in mind that these are three different loads but the approach is still the same. If you look at the mean +/- 2*SD then you'll notice that once again, 100% of the velocities easily fall within that range. By the way, I never use the SD reported by my chronograph (CED M2). I always import the data into Excel and use the =stdevpa() function. I personally don't think that "the properties of the normal distribution are very well understood" when it comes to bullet velocites and statistical implications for long-range shooting. When you consider that the better chronographs claim an accuracy of +/- 1% of the actual velocity, it makes me wonder if SD has any meaning in the real world as it pertains to ballistics. I will add that I'm here to learn and hope that I will be enlightened. I've put a lot of work into this for myself but I have more questions than answers. Using statistical methods without considering the real-world implications is pointless. Anyone can rattle off a a definition from Wikipedia and kid themselves that they know what's going on, but understanding what, how and why is the important thing. That's what I do in my job day in and day out ... why should this be any different?

http://128.171.62.162/hawthorn-engineering/thr/rem700_308win/range_targets/2010/02_14_10/sd_data.jpg

:)

OK ... then take all 30 shots fired in the .308 Win data set above. Here are the results shown below. Bare in mind that these are three different loads but the approach is still the same. If you look at the mean +/- 2*SD then you'll notice that once again, 100% of the velocities easily fall within that range. By the way, I never use the SD reported by my chronograph (CED M2). I always import the data into Excel and use the =stdevpa() function. I personally don't think that "the properties of the normal distribution are very well understood" when it comes to bullet velocites and statistical implications for long-range shooting. When you consider that the better chronographs claim an accuracy of +/- 1% of the actual velocity, it makes me wonder if SD has any meaning in the real world as it pertains to ballistics. I will add that I'm here to learn and hope that I will be enlightened. I've put a lot of work into this for myself but I have more questions than answers. Using statistical methods without considering the real-world implications is pointless. Anyone can rattle off a a definition from Wikipedia and kid themselves that they know what's going on, but understanding what, how and why is the important thing. That's what I do in my job day in and day out ... why should this be any different?

http://128.171.62.162/hawthorn-engineering/thr/rem700_308win/range_targets/2010/02_14_10/sd_data.jpg

:)

DickM

August 5, 2010, 05:37 PM

Well, let me start by saying that you couldn't be more wrong about the properties of normal distributions not being well understood, and it doesn't matter whether we're talking about distributions of bullet velocities, IQs, or sperm counts (just examples, you understand - I don't know what distributions those last two data types might follow, and I'd expect sperm counts, at least, to be lognormal) - a normal distribution is a normal distribution is a normal distribution and it doesn't matter where the data come from. We know the properties of normal distributions as absolutely as we know that 1+1=2. The properties of normal distributions, as well as those of several other distributions that data sets can follow, are the foundation of everything we do in statistics, which is in turn the foundation of everything we do in scientific research (which, since you brought it up, is what I do in my job day in and day out). (Climbing off soapbox now).

Now if what you're saying is that bullet velocities don't follow a normal distribution, then that's different. All of the chrono data I've looked at, including your data, do not deviate significantly from normal. That's not the same as saying they actually are normal, but usually after doing the appropriate tests and finding no reason to believe they're not normal (or whatever alternate distribution we think they might be) we proceed along assuming that they are normal. We'll be wrong from time to time, and the tests for normality tell us exactly how often we'll be wrong (but not exactly when, unfortunately). All I can say is that all the tests I've done indicate that normality is an appropriate assumption for populations of bullet velocities - I'll accept that that might be incorrect (but I honestly don't think so).

Now, with regard to the large data set from your last post, here's the problem. Each of the three individual 10-shot data sets appears to be normally distributed, as I would expect them to be. But notice that they have very different means, and when you combine them you get a distribution that has three areas of high density, clustered around each of the three means. That combined data set (which would be termed "trimodal") very definitely does not conform to a normal distribution (the test tells me that conclusion will be wrong something less than 1% of the time, so we can have reasonable confidence in it). So, it's still possible to go ahead and calculate a standard deviation, but all bets are off about what it's telling you and you just can't draw any conclusions from it. If you happen to have a 30-shot sample from a single loading (and shot at the same time, same gun, etc.) then there's a good chance that distribution will be normal, and I'd be happy to take a look at it.

Now if what you're saying is that bullet velocities don't follow a normal distribution, then that's different. All of the chrono data I've looked at, including your data, do not deviate significantly from normal. That's not the same as saying they actually are normal, but usually after doing the appropriate tests and finding no reason to believe they're not normal (or whatever alternate distribution we think they might be) we proceed along assuming that they are normal. We'll be wrong from time to time, and the tests for normality tell us exactly how often we'll be wrong (but not exactly when, unfortunately). All I can say is that all the tests I've done indicate that normality is an appropriate assumption for populations of bullet velocities - I'll accept that that might be incorrect (but I honestly don't think so).

Now, with regard to the large data set from your last post, here's the problem. Each of the three individual 10-shot data sets appears to be normally distributed, as I would expect them to be. But notice that they have very different means, and when you combine them you get a distribution that has three areas of high density, clustered around each of the three means. That combined data set (which would be termed "trimodal") very definitely does not conform to a normal distribution (the test tells me that conclusion will be wrong something less than 1% of the time, so we can have reasonable confidence in it). So, it's still possible to go ahead and calculate a standard deviation, but all bets are off about what it's telling you and you just can't draw any conclusions from it. If you happen to have a 30-shot sample from a single loading (and shot at the same time, same gun, etc.) then there's a good chance that distribution will be normal, and I'd be happy to take a look at it.

SourMash

August 5, 2010, 06:25 PM

Am I the only one that didn't take calculus or triganometry in high school? I barely passed the "easy" Algebra classes.

THe Dove

August 5, 2010, 06:44 PM

BWAHAHHAHAHAHAHHAHAHAHHAHA

You the man Sour Mash!!!!! I feel like an idiot too!!!!!! At least you can spell Calculus and Trig!!!!

The Dove

You the man Sour Mash!!!!! I feel like an idiot too!!!!!! At least you can spell Calculus and Trig!!!!

The Dove

MCMXI

August 5, 2010, 08:41 PM

Now if what you're saying is that bullet velocities don't follow a normal distribution, then that's different.

That's EXACTLY what I'm saying. You can force a Gaussian (normal) distribution on any data set (see below) and end up with a decent looking bell curve but does that mean the data is a normal distribution? Surely, bullet velocity is a function of numerous random variables measured by a less than perfect chronograph, the SD of which is a function of more random variables. My contention is that my velocity data doesn't fit the normal Gaussian model because bullet velocity isn't a normal distribution.

So here's a plot of my 30 shots (.308 Win) shown above. Looks like a classic Gaussian plot right ... trimodal if you like?

http://128.171.62.162/hawthorn-engineering/thr/ballistics/standard_deviation/normal_distribution.jpg

:)

That's EXACTLY what I'm saying. You can force a Gaussian (normal) distribution on any data set (see below) and end up with a decent looking bell curve but does that mean the data is a normal distribution? Surely, bullet velocity is a function of numerous random variables measured by a less than perfect chronograph, the SD of which is a function of more random variables. My contention is that my velocity data doesn't fit the normal Gaussian model because bullet velocity isn't a normal distribution.

So here's a plot of my 30 shots (.308 Win) shown above. Looks like a classic Gaussian plot right ... trimodal if you like?

http://128.171.62.162/hawthorn-engineering/thr/ballistics/standard_deviation/normal_distribution.jpg

:)

MCMXI

August 5, 2010, 08:57 PM

Well, let me start by saying that you couldn't be more wrong about the properties of normal distributions not being well understood

If you read what I said again you'll notice that I specifically stated ...

I personally don't think that "the properties of the normal distribution are very well understood" when it comes to bullet velocites and statistical implications for long-range shooting.

There may well be others that see things differently and I'm here to be enlightened so if you can better my understanding by relating statistical methods to real-world results and implications then I'm all ears.

:)

If you read what I said again you'll notice that I specifically stated ...

I personally don't think that "the properties of the normal distribution are very well understood" when it comes to bullet velocites and statistical implications for long-range shooting.

There may well be others that see things differently and I'm here to be enlightened so if you can better my understanding by relating statistical methods to real-world results and implications then I'm all ears.

:)

DickM

August 5, 2010, 11:04 PM

. . . I'm here to be enlightened so if you can better my understanding by relating statistical methods to real-world results and implications then I'm all ears.

I tried to do that, but you seem to only want to argue about it, attempting to support that argument with pseudostatistical gibberish. Adios.

I tried to do that, but you seem to only want to argue about it, attempting to support that argument with pseudostatistical gibberish. Adios.

MCMXI

August 5, 2010, 11:11 PM

Huh ... don't get all bent out of shape over this. Clearly you're not 100% sure based on your statement below. If you don't want to find out for sure then just say so.

Now if what you're saying is that bullet velocities don't follow a normal distribution, then that's different. All of the chrono data I've looked at, including your data, do not deviate significantly from normal. That's not the same as saying they actually are normal, but usually after doing the appropriate tests and finding no reason to believe they're not normal (or whatever alternate distribution we think they might be) we proceed along assuming that they are normal. We'll be wrong from time to time, and the tests for normality tell us exactly how often we'll be wrong (but not exactly when, unfortunately). All I can say is that all the tests I've done indicate that normality is an appropriate assumption for populations of bullet velocities - I'll accept that that might be incorrect (but I honestly don't think so).

:)

Now if what you're saying is that bullet velocities don't follow a normal distribution, then that's different. All of the chrono data I've looked at, including your data, do not deviate significantly from normal. That's not the same as saying they actually are normal, but usually after doing the appropriate tests and finding no reason to believe they're not normal (or whatever alternate distribution we think they might be) we proceed along assuming that they are normal. We'll be wrong from time to time, and the tests for normality tell us exactly how often we'll be wrong (but not exactly when, unfortunately). All I can say is that all the tests I've done indicate that normality is an appropriate assumption for populations of bullet velocities - I'll accept that that might be incorrect (but I honestly don't think so).

:)

MCMXI

August 5, 2010, 11:14 PM

I tried to do that, but you seem to only want to argue about it, attempting to support that argument with pseudostatistical gibberish. Adios.

That's a bit harsh!! I plotted my data set using the accepted normal function so what's the problem?

:)

That's a bit harsh!! I plotted my data set using the accepted normal function so what's the problem?

:)

MEHavey

August 6, 2010, 09:31 AM

Good grief, guys.

By the time the ambient temp and hunting altitude/humidity changes; you're using a different lot of primer and powder; AND bounce it off a few twigs getting to the the Muley (who never took geometry, much less trig).... only St Thomas Aquinas would be impressed.

:neener:

By the time the ambient temp and hunting altitude/humidity changes; you're using a different lot of primer and powder; AND bounce it off a few twigs getting to the the Muley (who never took geometry, much less trig).... only St Thomas Aquinas would be impressed.

:neener:

Sport45

August 6, 2010, 11:55 PM

Torque can be explained very quickly. In fact, it only takes a moment...

Good one! :)

Good one! :)

THe Dove

August 7, 2010, 02:47 PM

Does the torque explaination have all these graphs and stuff????? Cause if so, then I think it will take more than a second.

The Dove

The Dove

dmazur

August 7, 2010, 06:54 PM

Does the torque explaination have all these graphs and stuff????? Cause if so, then I think it will take more than a second.

From Wikipedia -

Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist. (emphasis added)

http://en.wikipedia.org/wiki/Torque

The attempt at humor was that "torque" and "moment" are sometimes used as synonyms, depending on whether you are dealing with a physicist or an engineer. This was thin, at best, and had nothing to do with the amount of time required for an explanation.

Most folks are familiar with "torque specifications", or at least have heard of tightening a nut or bolt to x ft-lbs. They may nor may not be able to determine how many lbs of force are required on an 18" wrench to generate 75 ft-lbs of torque on a lug nut. (Answer: 50 lbs)

This is buried in the Wikepedia article under Special Cases and Other Facts, Moment Arm Formula, and it's really all most of us need to know about torque.

So, that "special case" is an explanation of torque, without graphs and stuff. :)

From Wikipedia -

Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist. (emphasis added)

http://en.wikipedia.org/wiki/Torque

The attempt at humor was that "torque" and "moment" are sometimes used as synonyms, depending on whether you are dealing with a physicist or an engineer. This was thin, at best, and had nothing to do with the amount of time required for an explanation.

Most folks are familiar with "torque specifications", or at least have heard of tightening a nut or bolt to x ft-lbs. They may nor may not be able to determine how many lbs of force are required on an 18" wrench to generate 75 ft-lbs of torque on a lug nut. (Answer: 50 lbs)

This is buried in the Wikepedia article under Special Cases and Other Facts, Moment Arm Formula, and it's really all most of us need to know about torque.

So, that "special case" is an explanation of torque, without graphs and stuff. :)

NoAlibi

August 7, 2010, 07:53 PM

Mathematician Wars on THR - imagine that! :what:

Well, who ever is full of xxxx can always work it out with a slide rule! :neener:

PS - dmazur: "...it only takes a moment" - clever fellow, too funny! :D

Well, who ever is full of xxxx can always work it out with a slide rule! :neener:

PS - dmazur: "...it only takes a moment" - clever fellow, too funny! :D

THe Dove

August 7, 2010, 08:16 PM

I just wish I could get a graph or something for my question!! ;-(

The Dove

The Dove

MEHavey

August 7, 2010, 10:59 PM

What do you want to know about torque?

Shadow 7D

August 8, 2010, 07:46 AM

Not much, that would be why he is asking

Actually, mathematically many forms of distribution can be forced onto a data set, but until you look at the data and analyze it, finding your population, whether continuous or not, you don't come to the correct way of normalizing your distribution, I have a advanced college statistics book if anybody wants it, used twice and poorly understood.

Actually, mathematically many forms of distribution can be forced onto a data set, but until you look at the data and analyze it, finding your population, whether continuous or not, you don't come to the correct way of normalizing your distribution, I have a advanced college statistics book if anybody wants it, used twice and poorly understood.

MEHavey

August 8, 2010, 06:16 PM

"Statistics" and "Torque" are apples & oranges.

They have nothing to do with each other.

That said,....

If I hold a 1-lb weight in my hand, that weight causes 1-lb of Force straight down on my hand.

If I have a wrench 1-ft long applied to a bolt, and I put that same 1-lb weight on the end of it, the bolt is twisted with "1 Ft-lb" of TWISTing force called Torque, or "Moment"

If I put 2-lbs on the end of that 1-ft wrench, I get 2 ft-lbs of torque

If I put 1-lb weight on a two-foot wrench, I (again) get 2 ft-lbs of torque

If I put a 10-lb weight on that same 2-ft wrench I get 20 ft-lbs of torque

Torque is twisting force

The more the force on the end of a bar, the more the torque.

The longer the bar, the more the torque.

So simple even a caveman can do it. :)

They have nothing to do with each other.

That said,....

If I hold a 1-lb weight in my hand, that weight causes 1-lb of Force straight down on my hand.

If I have a wrench 1-ft long applied to a bolt, and I put that same 1-lb weight on the end of it, the bolt is twisted with "1 Ft-lb" of TWISTing force called Torque, or "Moment"

If I put 2-lbs on the end of that 1-ft wrench, I get 2 ft-lbs of torque

If I put 1-lb weight on a two-foot wrench, I (again) get 2 ft-lbs of torque

If I put a 10-lb weight on that same 2-ft wrench I get 20 ft-lbs of torque

Torque is twisting force

The more the force on the end of a bar, the more the torque.

The longer the bar, the more the torque.

So simple even a caveman can do it. :)

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