Standards exist for two main reasons. Say that you want to build a fairly permanent shelter. If you want the shelter to be sturdy, the parts will have to relate fairly accurately. Some of the members will need to meet at right angles; some of them will need to be of equal length. Of course you can figure out which ones need to be of equal length, cut one, and then use that one as a standard to cut the rest – that would be a temporary standard. But if you use that standard in the construction, it won’t be available for the next thing you want to build. You could save a board of standard size but a board would be difficult to carry around. Perhaps a walking staff of standard length would be better.
The other reason for standards is so that you can communicate a procedure to others. Say that there are three people working on the shelter. You could cut a board and tell someone else, “Cut three more boards the same length as this one.” But, what if you run into another party on an expedition and they want to know how to build the shelter? You have no standard with you. You could say, “Cut a board the length of three of these walking staffs.” Then you would have to give them your walking staff or cut another stick the size of your walking staff. In that case, a new permanent standard is developing and being transmitted.
How much easier would it be to say, “Cut four boards nine feet long”? The foot is a universally understood standard. So the question becomes, “How could you easily determine a foot (or any other universally understood standard) in a survival situation when you don’t have a foot ruler, yardstick, or tape measure to refer to?”
With some preparation, the solution is not difficult. Let’s look at some possibilities.
Fortunately, Uncle Sam is obsessive about standards. All coins of the same denomination are pretty much the same size. I had four quarters, 3 nickels, 5 pennies, and 5 dimes to work with and I measured the diameter of each 5 times, turning them so that I would get readings of their diameters in several places. The readings did not differ by more than 1/128″ on any coin in any direction. The results were as follows:
The diameter of the quarters were 15/16″ + 3/256″ ± 1/256″.
The diameter of the nickels were 13/16″ + 7/256″ ± 1/256″.
The diameter of the pennies were 3/4″.
The diameter of the dimes were 11/16′ + 5/256″ ± 1/256″.
I noticed that none of the coins varied by more than 1/128″. It’s very rare in a survival situation that you will need this kind of precision. But the dimensions of most of the coins are not particularly useful as standards. For example, a quarter is very close to 15/16″. One quarter is pretty close to an inch but if you use 12 quarters to approximate a foot, you will be off by 12/16″. That’s 3/4′ and that will be too inaccurate for even survival situations.
One coin is very convenient, though, and it’s normally a nuisance – I hate being stuck with a pocket full of pennies. A penny is almost exactly 3/4′. That means that 4 pennies lined up is 3 inches long. To do that, set up a straight edge (a T-square would be ideal since you could set the first penny against the cross member of the T-square). Since the pennies are all of equal diameter, the upper edges will make a parallel line with the straight edge. If you line up 4 pennies like this, the left edge of the first penny on the left would be almost exactly 3 inches from the right edge of the last penny at the right of the line up. Also, if you line up 16 pennies like this, you will have a standard foot.
There are other ready standards for length. Some are more useful than others. Most architectural materials are made in standard sizes but it’s hard to use a two-by-four as a straight edge. On the other hand, most floor tiles are one foot square and can very easily be used as a straightedge.
A very ready standard is your own body. If you were a Boy or Girl Scout, you probably developed standards for use outdoor by measuring different parts of your body. The most common body standards are your stride and the distance between knuckle joints.
To use a finger you have to find a finger section that is close to an inch and, obviously, that’s going to only be an approximate measure. It turns out that the middle joint on my right little finger is very close to an inch in length.
Stride length can be much more useful for larger distances. To find your stride length, set out a 200 foot path and walk it, finding the number of steps you take. Do that 10 times and take the average number of steps. I performed this simple exercise at the 2005 Summer SEHowl and came up with the following numbers of steps per 20 feet.
80, 75, 81, 79, 82, 76, 79, 80, 81, 78
The average was then the sum of the steps (791) divided by the number of trials (10) or 79.1 steps per 20 feet. Most people’s stride will be close to 80 steps or 100 steps per 200 feet.
If the stride is 80 steps per 200 feet, one can step of a distance and multiply their number of steps by 2.5 to determine the number of feet in the distance.
If the stride is 100 steps per 200 feet, one multiplies their number of steps by 2 to get the number if feet.
My average stride is close enough to 80 to be fairly accurate using 2.5 as the multiplier to get distance in feet by stepping off a distance.
The average number of steps gives an idea of what my typical stride is like but it’s interesting to know how much my stride varies. There’s a value that will tell me that; it’s called the standard deviation. It’s calculated by the following procedure.
Add all the values. (791)
Square all the values and add all the squared values.
(62613)
Square the sum obtained in step 1 and divide this value by the number of values that were added to obtain this sum. (62568.1)
Subtract the value obtained in step 4 from the sum in step 2. (44.9)
Divide the last value by the number of repetitions minus 1. (4.99)
Take the square root of this last value. (2.23)
In a “normal case” (and I’ll talk about that in a minute), 68.2 % of the time, any randomly selected value will be within 2.23 steps from the mean; and 95.4% of the time it will be with in 4.46 steps from the mean. Therefore, roughly 68% of the time any time I walk 20 feet, I will have taken between 79.1 – 2.23
= 76.9 and 79.1 + 2.23 = 81.3 steps. Roughly 95% of the time, I will have taken between 74.6 and 83.6 steps.
That’s not too bad. 68 % of the time I will only be off at the most by 2.23 steps which is only 5.57 feet out of 200 feet.
There’s another number that expresses the amount of error in my stride. The standard error of measurement tells me how far off my stride can be expected to be 68% of the time. It’s calculated by dividing the standard deviation by the square root of he number of measurements. In this case, that would be 2.23 divided by the square root of 10 or approximately .7 steps. In other words, I can be fairly sure that I will be within give or take .7 steps of the true distance when I step off a distance like this. That’s only 1.6 feet. If I step off 100 feet, the true measurement should be between 98.4 and 101.6 feet.
Before I go on to talk about what “normality” means, I would like to point out the problem with pacing off heel-to-toe. It’s simply that it depends on which shoes you’re wearing. Shoe size relates to the length of your foot but different shoes allow you more or less room ahead of your toes. Therefore, heel-to-toe distance will vary greatly with which shows you’re wearing.
Here are a few experiments that, although they are very different in their mechanics, will yield very similar results:
1. In a column, write the numerals 2 through 12. Throw a couple of dice 100 times. For each throw, make a tally beside the number that comes up. When I did that I came up with the following results:
Throw Tallies Number of tallies
2 ||| 3
3 |||| 4
4 |||||||||| 10
5 |||||||| 8
6 |||||||||||||||| 16
7 |||||||||||||||| 16
8 |||||||||||||||||| 18
9 |||||| 6
10 ||||||||| 9
11 |||||| 6
12 |||| 4
100
These results aren’t particularly surprising. In throwing two dice, there’s only one way for a 2 to come up – a 1 on each die. To get a 3 you have to throw a 1 on one dice and a two on the other. But there are two ways to get a 4. You can either throw a 1 and a 3 or you can throw two 2s. There are three ways to get a 7: 1 + 6, 2 + 5, 3 + 4. So you would expect to come up with more 4s than 2s or 3s, and more 6s, 7s, and 8s than 4s or 5s. Notice the shape of the tallies. You’ll see the shape again (and again…..)
2. Throw one die 10 times and then average the values (add them and divide by 10. Round that value to the nearest whole number. Repeat 100 times recording each result. Finally make a tally of all the averages that came up as 1, 2, 3, 4, 5, or 6. My results looked like this:
Average Tallies Number of tallies
1 0
2 ||| 3
3 |||||||||||||||||||||||||||||||||| 34
4 |||||||||||||||||||||||||||||||||||||||||||||||||||||| 54
5 ||||||||| 9
6 0
There’s that pattern again. How did it happen this time? Well, that’s sorta complicated so I will just give you the short version and let you look it up if you wanna. I just want to show you that it works, here. Mathematicians call it the Central Limit Theorem and it states that if you take the sum of many identically distributed independent random variables with finite variance, and you do it again and again, charting the results, a curve like the ones above will result. Dice throws are identically distributed – each possible outcome (1,2,3,4,5,6) has the same probability of coming up (1 out of 6) in any throw. They are independent because what you throw on one die doesn’t affect what will come up on another die or another throw of the same die. And the throw of a die can only vary between 1 and 6 so the variance is finite. This result is very important since a large amount of statistical science is based on it.
The curve is called the Normal Curve and the experimental results above are not perfect representations – there’s some error in the processes since incidental things such as drafts and dice throwing techniques remove some of the randomness of the processes. The ideal normal curve is symmetric. In other words, if you drew a perfect normal curve and folded it in half, both halves would lie exactly together.
Let’s, look at another way to get a normal curve.
Take a pegboard and place rows of staggered pegs in the holes (as shown below):
O O O O O O
O O O O O
O O O O O O
O O O O O
The bottom of the board should be set up so that it will catch marbles that are dropped into the center of the board at the top. The marbles should be just large enough to pass through the pegs. If you drop several marbles into the top of the board, you will see them collect at the bottom in a normal curve. Bob Barker has a variation of this called Plinko on The Price is Right. You can play with the results by using smaller or differently shaped objects than marbles or by dropping them at different places on the board. I used quarters (see the picture above).
If you don’t have a pegboard, you can simulate a Galton Board (that’s what the device is called) by tossing coins. Notice that, the reason the board works is that at each level, the marble has two options – it can go right or left around a peg. So to simulate a Galton board with six levels of pegs, toss a coin six times. If the coin lands with heads up, add a one to the result to simulate a marble going right. If the coin lands tails up, subtract a one to simulate a marble going left. In the end you’ll have a number between –6 and 6. To simulate a board with 100 marbles, you would repeat the above procedure 100 times. When I did this, I came up with the following results:
Result of 6 tosses Tallies Number of tallies
-6 0
-5 0
-4 ||||||||||||| 13
-3 0
-2 |||||||||||||||||||||||||||||| 30
-1 0
0 ||||||||||||||||||||||||||||||| 31
1 0
2 ||||||||||||||||||||| 21
3 0
4 ||||| 5
5 0
6 | 1
Well, this makes sense like the two dice experiment. For a marble to reach slot –6, it could only do so by always going left around pegs. On the other hand, there are several ways for it to get into the 0 slot. One is left-right-left-right-left-right. Another is left-left-left-right-right-right. So it makes sense that most of the marbles will end up in the slot with the largest number of ways to end up there.
Many natural processes are normal in character. If you perform the pacing experiment over and over, you will find that it is no exception. Even the few repetitions in my experiment shows the beginnings of a normal curve:
Paces
75 |
76 |
77
78
79 |||
80 ||
81 ||
82 |
So, assuming the normality of pacing, I can derive quite a lot of information about measuring distance using my own stride.
We’re going to come back to length later when we consider how to make a usable standard instrument to measure length – a ruler – and even later when we consider how to measure inaccessible distances by using surveying techniques but, for now, let’s stay with the basics.
There’s another important quantity that we often need to measure in survival situations – weight. Different concoctions such as soaps, dyes, and medicinals can be rather unforgiving of sloppy compounding so we need to know how much of each chemical we’re mixing together if we want to have any hope of coming out with a usable product. In this case we need to have a standard way of measuring weight before we can come out with some standards of weight. Fortunately, that’s easy.
There are actually two ways of measuring weight – and I just told a small lie. So let me explain the problem of what I just said. There are two things that are often confused (in fact, they are often confused intentionally for the sake of convenience). There’s mass and there’s weight. Mass is a quality of matter. It’s a measure of how much matter there is in a body. A one gram body has one gram of matter in it. Weight is the force exerted by an object toward the earth due to gravity: technically it’s equal to the mass of the object multiplied by the local gravitational acceleration. Grams and pounds are, technically units of mass. The English Stone is a unit of weight.
You measure mass by comparing an object with another object of known mass. You measure weight by comparing the downward force of a hanging object with a known force. Let’s look at how you measure weight first and then we’ll forget it and the distinction between weight and mass. In terms of survival technology, the distinction is quite forgettable.
The most common means of measuring weight is the spring scale. You use it in the supermarket to measure vegetables and, if you’re a fisherman, you use it to weight your catch. A spring scale pits an unknown weight against the force exerted by a spring that exerts a known force. The problem with springs (and rubber bands, and other springy things) is that they become more or less springy over time and have to be recalibrated occasionally. Spring scales are useful in a lot of situations and I have several in my gadget kit, but we’re going to start ignoring them now.
Mass is measured by the balance. A balance is simply an arrangement that balanced (thus the name) one object against another.
I was in Fred’s the other day and saw some nicely made wooden rulers with metal inserts along each edge. They were ridiculously cheap so I bought three. They have a hole at six inches and a hole about one and three quarters inches from each end. I looked through them and found three that were fairly accurate.
In constructing my balance, I was pretty sloppy because I already have a little portable gold assayers balance that I bought for next to nothing. I’ll be using that to play with mass later on, but, if you repeat my experiments here and want to make a permanent balance you may want to be a little more precise. For instance, I’m using paper clips for a rider but you can cut a very practical metal slider out of an aluminum can.
I tied the ruler by a thread to the cross bar of my television cart (which is what I use for a laboratory table in my apartment. Then I cut the bottoms out of two paper cups to use as the two trays of the balance. I used a needle to run a thread through three points in the bottom of each cup bottom and tied each so that I could hang them from paper clips in the end holes in the ruler.
The result, of course, didn’t balance. There are several reasons – all predictable. The wood used to make the ruler isn’t perfectly homogeneous, the holes are not exactly placed, the cup bottoms were roughly cut out of the cups with an Exacto knife, the paper clips are not exactly he same sizes, etc., etc. The nice thing about a balance is that it’s easy to fix all that.
Surprisingly, despite my sloppy workmanship, my balance actually hung at about 110° to the vertical – almost balanced. To remedy that, I placed a paperclip over the upper edge of the ruler at 6 5/8 inches to bring the whole thing into balance. The paperclip is what I called above the “rider”. It’s called that because it rides on the beam.
Note, all that’s necessary to have a usable balance is to have a cross beam with cups hanging from each end – that balances. Now, at present, my balance requires me to judge if it is balanced (that the beam is horizontal) by simply looking at it. For really precise measurements, that’s not really enough. For a more precise instrument, you might want to have a plumb bob hanging in front of the 6 inch mark so that you can judge if if coincides with a mark on the ruler that’s at 90° to each edge. There’s quite a few ways to judge if the beam is horizontal. My assayer’s balance has a pointer and scale that indicates when the beam is balanced.
There’s several ways to do what I’m about to do. I happen to have a couple of sets of standard masses (one is left over from a 4+ year stint in pharmacy school – it’s a beauty). The other is smaller and came with my assayer’s balance. I’m going to check the characteristic of the rider on the balance. That will give a good idea of the relationship of it’s distance from the balance point (the 6 inch mark) and the weight in the opposite cup, and along the way I’ll introduce you to regression analysis.
You don’t have to use standard weights. All you need is small weights that are pretty much the same. BBs would be perfect. You can count the number of BBs required to make the beam balance Beads of constant size is another possibility. We don’t have to be exact here.
What I’m going to do is move the rider (paperclip) by one half inch increments from the balance point to the end of the beam and record how much weight is required in the opposite cup to balance the beam. then we’ll play with the numbers.
Okay, here’s my data:
When the rider was at – it took this many milligrams of mass to balance it
7 – 50
7.5 – 200
8 – 250
8.5 – 300
9 – 350
9.5 – 400
10 – 450
10.5 – 500
11 – 550
11.5 – 600
12 – 650
Obviously, the further the rider was from the balance point of the beam, the more mass it took to balance it. Also, if you did this little experiment, you might have noticed that it did matter where you placed the weight on the cup. That makes sense. my cups were abut three inches in diameter so a weight placed in the cup could have been as much as three inches closer or further away from the beam’s balance point.
If you graph these points, placing the distance of the rider from the balance point on the x axis of the chart and the masses on the y axis, you’ll notice that you get a straight line except for the very first of the curve at 7 inches.
When you have a straight line on a graph, you say that the two quantities being graphed have a linear relationship. When an increase in one is accompanied by an increase in the other, you say that the relationship is direct. If the increase in one is accompanied by a decrease in the other, you say that the relationship is indirect.
For my ruler balance, the relationship of the distance of the paperclip rider from the balance point of the ruler beam to the amount of mass required to balance it is direct and linear from 7.5 inches to 1 inches. There is a nonlinearity when the rider is less than 7.5 inches from the balance point. Since the rider balances the beam when it’s at 6 5/8 inches, I sorta expected something to go “wrong” there. It’s not that the laws of physics is out of whack near a balance point, it’s just that a lot of things can go wrong in procedure and design there.
In the linear region of my beam, I can say that for every half inch away from the balance point I place my rider, I need 50 more milligrams to balance the beam. I can express that by an equation called a regression equation. If d is distance from the balance point (in inches) and m is the mass required to balance the beam (in milligrams), then:
m = 100d – 550
If you’ve ever had analytical geometry, you’ll know that there’s several equations for a straight line, but let me only go over one here. You know, from the data above, that it takes 50 milligrams to balance the beam for every move of one half inch of the rider away from the balance point. On the graph, that means that for every 50 milligrams up you move, the line moves .5 to the right. The slope of the line is, therefore, 50/.5 or 100. If you extend the line backward, you see that it would cross the x axis at m = -550. If you start at d = 0 and m = -550 and start moving 1 to the right and 100 up, you will eventually reach our line. So, to find the mass needed to balance the beam, you multiply the distance of the rider from the balance point by the slope of the line on the graph and subtract 550.
This data is rather nice with little error in it. There are statistical equations that can be used to find the best line through a bunch of messy data points that don’t follow a straight line but do cluster around a straight line. Finding the equation for such a line is called linear regression analysis. And if you want to know more about it right now, you can easily look it up. We don’t need to get that advanced yet. Most physical data is pretty well behaved and you can pretty much take a ruler and draw the closest straight line through data points.
Now. I’m going to switch over to my assayer’s balance to see if I can find some standard weights among common objects.
An obvious starting place (remembering Uncle Sam’s obsession with consistency) is coins. When I was a kid with a chemistry set, I learned that, within a very satisfactory degree of accuracy, a dime weighed 2 1/2 grams, a nickel weighed 5 grams, a penny weighed 3.1 grams, and a silver dollar weighed 12 1/2 grams. But that was back when coins were not made of mystery metal. I have no idea what they weigh now.
I took ten coins each – dimes, pennies, nickels, and quarters (dollar coins are too rare any more and too irregular to be considered as a convenient standard) and weighed them. The results are as follows:
Dimes: Nine of the dimes were between 2.1 and 2.2 grams. Unfortunately, the other dime was 2.5 grams. Still, the average was 2.2 grams with a standard deviation of 0.11 gram and a standard error of 0.04 gram, which is statistically pretty good. In other words, there may be some outliers but dimes, as a group tend to be pretty reliable. My recommendation – if one dime out of ten is drastically different, it would be dangerous to put much stock in the weight of a single dime; but it is much more likely that, for instance, ten dimes will sum to 22 grams. If you balance 10 dimes with a square of material cut out of an aluminum can, you can be fairly certain that it will be acceptably close (for survival purposes) to 22 grams.
Pennies: Again, nine of the pennies were very similar in mass, ranging from 2.45 to 2.6 grams. But one was 3 grams. So there’s the same problem. The average was 2.6 grams with a standard deviation of 0.16 and a standard error of 0.05 gram. Again, combining ten pennies would give you a fairly accurate 26 gram weight.
Nickels: The nickels were more consistent, ranging from 4.8 to 5 grams. The average is 4.9 grams with a standard deviation of 0.12 and a standard error of 0.04. The problem is that my sample is small and, given the unreliability of the other two denominations, I can’t be too sure that an eleventh nickel won’t be, say, 5.5 grams. Still, ten nickels should be pretty close to 50 grams and that’s a nice number. If you counterbalanced ten nickels with five pieces of aluminum of equal size, you would have five 10 gram masses. You might be tempted to use one of the 10 gram masses to come up with two 5 gram masses or ten 1 gram masses but remember that each time you subdivide the weight, you are going to be introducing a small but significant amount of error. These errors will add up quickly with subsequent divisions.
Quarters: The quarters were also surprisingly consistent (given the variety of designs of both the nickels and the quarters over the past 20 years). They ranged from 5.45 to 5.7 grams with an average mass of 5.5 grams, a standard deviation of 0.08 and a standard error of 0.02 grams. Again, ten quarters should counterbalance a 55 gram mass fairly accurately.
The Therian Timeline 