187. Space objects evenly by adding “1”

[Make Things]

[mklotz]

The " + 1 " appears automatically if you think about how many spaces you need to create.

If you arrange N screws in a row you automatically create N - 1 spaces. Two screws create one space between them, three screws create two spaces, four create three, and so on. However you need to add the two spaces at the ends, one between the leftmost screw and the leftmost edge of the board, one between the rightmost screw and the rightmost edge of the board. Thus, you need

N - 1 + 2 = N + 1

spaces and voilà, there's your " + 1 ", and dividing N + 1 into the length of the board will give you the size of the required space(s).

Note that this formulation works for the case of only one screw, yielding a space equal to half the width of the board, i.e. put the screw in the middle of the board.

It also works for zero screws, yielding a spacing equal to the board width.

While the latter two cases aren't of practical interest, they're mathematically satisfying showing the formulation would work for all possible values of N.

[WmRMeyers]

The " + 1 " appears automatically if you think about how many spaces you need to create.

If you arrange N screws in a row you automatically create N - 1 spaces. Two screws create one space between them, three screws create two spaces, four create three, and so on. However you need to add the two spaces at the ends, one between the leftmost screw and the leftmost edge of the board, one between the rightmost screw and the rightmost edge of the board. Thus, you need

N - 1 + 2 = N + 1

spaces and voilà, there's your " + 1 ", and dividing N + 1 into the length of the board will give you the size of the required space(s).

Note that this formulation works for the case of only one screw, yielding a space equal to half the width of the board, i.e. put the screw in the middle of the board.

It also works for zero screws, yielding a spacing equal to the board width.

While the latter two cases aren't of practical interest, they're mathematically satisfying showing the formulation would work for all possible values of N.

It's called a "fence post problem" in some circles. I've got parts of a book case here that I made many years ago before I learned this. Plan was to make folding book cases for standard paper-back books, of which I had thousands. 6' tall by 2' wide, made from 1x5 lumber, IIRC. Wound up with 6 perfect shelves, and one very short shelf... They were to be backed with 1/4" plywood, and hinged so they closed to form a box around the books. I was gonna do a whole bunch of them! That idea crashed and burned on my very limited math skills. (Hah!) Haven't gotten around to it since them, either. That was in preparation for my 36th move, I think. I'm around 70 or 80 moves, now. And the book herd has been thinned a few times, by floods, children, and natural wear and tear on cheap paperbacks. Though I've been in this place for 27 years middle of next month. Before that, longest I'd ever lived in one place was 5 years, and that was 3 different residences.

[imohtep56]

It's called a "fence post problem" in some circles. I've got parts of a book case here that I made many years ago before I learned this. Plan was to make folding book cases for standard paper-back books, of which I had thousands. 6' tall by 2' wide, made from 1x5 lumber, IIRC. Wound up with 6 perfect shelves, and one very short shelf... They were to be backed with 1/4" plywood, and hinged so they closed to form a box around the books. I was gonna do a whole bunch of them! That idea crashed and burned on my very limited math skills. (Hah!) Haven't gotten around to it since them, either. That was in preparation for my 36th move, I think. I'm around 70 or 80 moves, now. And the book herd has been thinned a few times, by floods, children, and natural wear and tear on cheap paperbacks. Though I've been in this place for 27 years middle of next month. Before that, longest I'd ever lived in one place was 5 years, and that was 3 different residences.

Just take that very small shelf and add a stained glass panel to the front and backlight it...

[Make Things]

The " + 1 " appears automatically if you think about how many spaces you need to create.

If you arrange N screws in a row you automatically create N - 1 spaces. Two screws create one space between them, three screws create two spaces, four create three, and so on. However you need to add the two spaces at the ends, one between the leftmost screw and the leftmost edge of the board, one between the rightmost screw and the rightmost edge of the board. Thus, you need

N - 1 + 2 = N + 1

spaces and voilà, there's your " + 1 ", and dividing N + 1 into the length of the board will give you the size of the required space(s).

Note that this formulation works for the case of only one screw, yielding a space equal to half the width of the board, i.e. put the screw in the middle of the board.

It also works for zero screws, yielding a spacing equal to the board width.

While the latter two cases aren't of practical interest, they're mathematically satisfying showing the formulation would work for all possible values of N.

I knew there was a good formula to explain it...heck I searched for a method for a long time (many, many years ago). I stumbled across the idea by accident once and was so happy it worked I didn't think much of it but didn't want to jinx myself by thinking I'd finally solved the problem...kind of like, if I look into this too deep, I'll find that my return wasn't as perfect as I thought it would be.

I'm not even kidding...most of these little tips are from trial and error, and if I'm being honest, just dumb, stupid luck. Tips become fun when there's a problem that you find a solution to, and this one, literally took me 10 years to really peg down. I am happy to see there's an even better way to explain this than me just getting lucky with the solution.

[mklotz]

A large part of mathematical expertise is being able to see the problem at hand from different perspectives. Gauss, one of the three greatest mathematicians of the modern era, provided an excellent example of that skill.

The story is a tale from when Gauss was still at primary school. One day Gauss' teacher asked his class to add all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050.

Later, Gauss generalized this approach with the formula we still use today for the sum of the numbers from 1 to N, N*(N + 1) / 2.

[mklotz]

A large part of mathematical expertise is being able to see the problem at hand from different perspectives. Gauss, one of the three greatest mathematicians of the modern era, provided an excellent example of that skill.

The story is a tale from when Gauss was still at primary school. One day Gauss' teacher asked his class to add all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050.

Later, Gauss generalized this approach with the formula we still use today for the sum of the numbers from 1 to N, N*(N + 1) / 2.

Another, related, example of Gauss' lateral thinking involves a formula for the sum of the numbers between n1 and n2 (where n1 > 1 and n2 > n1).

Now you can derive this formula using the above expression for the sum from 1 to n, although on your first try you'll probably make a very common mistake.

Gauss noted that the formula for a common average...

A = S / N

where:

A = average
S = sum of items
N = number of items

could be solved for the sum

S = A * N

and A and N are easily computed...

A = (n1 + n2) / 2
N = (n2 - n1 +1)

so:

S = (n1 + n2) * (n2 - n1 +1) / 2

which is identical to what you would get using the expression for the sum from 1 to n

[Make Things]

Another, related, example of Gauss' lateral thinking involves a formula for the sum of the numbers between n1 and n2 (where n1 > 1 and n2 > n1).

Now you can derive this formula using the above expression for the sum from 1 to n, although on your first try you'll probably make a very common mistake.

Gauss noted that the formula for a common average...

A = S / N

where:

A = average
S = sum of items
N = number of items

could be solved for the sum

S = A * N

and A and N are easily computed...

A = (n1 + n2) / 2
N = (n2 - n1 +1)

so:

S = (n1 + n2) * (n2 - n1 +1) / 2

which is identical to what you would get using the expression for the sum from 1 to n

It's funny, I can do physics like object moving in my head, as I stare into the corner of the room. I can know that something I build will work before I cut the first piece (I think they call it spatial mechanical thinking) but math is such a quagmire of numbers and letters that make absolutely no sense to me, no matter how hard I try to arrange things or how many times I'm told how it works. It's incredibly frustrating. I spent 40 years of my life thinking I was an idiot because I barely passed pre algebra, yet I can build things that will work 99% of the time.

[sossol]

When I want to evenly space screws or nails in a constructed thing I mark for the first screw/nail at each end, then divide the distance between those marks instead of the entire span.

Neil

[mklotz]

When I want to evenly space screws or nails in a constructed thing I mark for the first screw/nail at each end, then divide the distance between those marks instead of the entire span.

Neil

And, if you're going to put N screws between the first two, and the distance between the first two is A, you calculate the spacing, S, as...

S = A / (N + 1)

because the N screws are going to create N + 1 spaces between the first two screws.