Workshop garrotte

[mklotz]

There are innumerable designs for tools to hold screws in place until you can get them started but the one I seem to use the most is one of my own design.  I call it a garrotte for obvious reasons and it's simple to fabricate in the home shop.

Mine consists of a loop of waxed, heavy-duty thread pulled through a length of 1/8" OD brass tubing.
A wire passing through the loop and anchored to the end of the tube prevents the garrotte loop from being pulled back into the tube when the ends projecting from the other end of the tube are pulled.

In use, it's looped around the screw to be positioned...



and pulled tight, thus trapping the screw.



The screw is still free to rotate about its axis.  The tube is then used to position the screw while a driver seats it into whatever receives the screw.

Takes about ten minutes to make and that includes waxing the thread.

I discovered it has a secondary use.  Until they perfect non-ferrous magnets, it makes a pretty good lasso for snaring small items dropped into finger-inaccessible holes.

[Tennessee Whiskey]

S snake snatcher for screws. I love it
E

[ths]

Another good one Marv. Love the birthday formula casually dropped into frame.

Hugh.

[DICKEYBIRD]

Could've used that one yesterday.  Thanks, will have to make one soon!

[mklotz]

Love the birthday formula casually dropped into frame.

Not quite.  That's part of the solution to a puzzle I posed to a friend of mine.  He had the nerve to give me a numerical answer that he obtained with some (spit) CAD program [one of the main reasons so many people can't do any math].  He acted as if I wanted the answer.  In much of math, the answer is irrelevant; what is important is the analysis to obtain it.

[b.lindsey]

Ingenious as always Marv....thanks!!!

Bill

[Pete49]

Brilliant idea Marv and was needed yesterday so a timely post. I wonder if I can scale it up and make it suitable to capture and restrain politicians??
Pete

[tel]

  I made several larger versions back when I was wrangling snakes and other critturs for a living. Never ever occurred to me to scale it down for workshop use. Suppose that just goes to prove you can't be have brains AND be good looking!

[Maryak]

Based on the mug shots so far on offer and on the comments below............................

Best Regards
Bob

[Jo]

 Thanks Tel: What does that say for me.. the truth is out   

Jo

[Alan Haisley]

Love the birthday formula casually dropped into frame.

Not quite.  That's part of the solution to a puzzle I posed to a friend of mine.  He had the nerve to give me a numerical answer that he obtained with some (spit) CAD program [one of the main reasons so many people can't do any math].  He acted as if I wanted the answer.  In much of math, the answer is irrelevant; what is important is the analysis to obtain it.

Ok, Marv, you whetted my curiosity.    Now I'm going to have to lurk in chatterbox to see if any puzzles show up.
(While the answer is sometimes relevant, the analysis is certainly where the fun is to be found.
Alan

[Pete49]

Jo rest assured it only applies to Tel....Having seen a photo of you a while back all I can say is the brain is packed in a beautiful package proving once and for all that brains and beauty can go together. Ignore Tel's comments as he is  just an old coot that's been caught out Maybe he should just 
Pete

[Jo]

  Thanks Pete you are a real gent 

But the photo you refer to was many  some a few a couple of years ago, I am still working on preventing any modern photo's appearing that might spoil your image of me 

Jo

[tel]

Watch it Jo - he probably wants to borrow something.

[mklotz]

Ok, Marv, you whetted my curiosity.    Now I'm going to have to lurk in chatterbox to see if any puzzles show up.
(While the answer is sometimes relevant, the analysis is certainly where the fun is to be found.

OK, here's the problem I posed...

Imagine a coordinate system scribed on a vertical wall.  One end of a 30 cm frictionless string is secured at the origin (0,0).  The other end is secured at (15,-12).  Free to slide on this string is a small weight.  What are the coordinates of the weight when it comes to rest?

Remember, no CAD solutions or experimental results need apply.  We want to see the problem worked out in mathematical form leading to expressions for the coordinates.

If that one frustrates you, here's an easier one I posed to him a while back.

One of the most famous equations in mathematics is:

e^(i*pi) = -1

where:

e = base of natural logarithms = 2.71828...
i = square root of -1
pi = 3.14159...

Prove that this is true. 

[Alan Haisley]

...
One of the most famous equations in mathematics is:

e^(i*pi) = -1
...
Prove that this is true.

To darn much stuff online. It is fairly easy to get to Euler's Identity, from there to Euler's Formula, and from there to several proofs.
The first one is more interesting in that I need to actually work it out (somehow), rather than just look up a solution and so it's the one I'm thinking about.
I remember sitting in class while the professor put a proof of Euler's Identity on the board, but I certainly don't remember it. Still, in terms of mathematical elegance I would say that it certainly beats out anything else.
I have to thank Vi Hart for one of her recreational mathematics videos in the Kahn Academy for reminding me where to search. I try to review her exposition on pi and tau every pi day; have to admit though that I haven't celebrated tau day yet.
Alan

[Tennessee Whiskey]

I hear bongos in the desert. You really do respect the mans work, cheers my friend.

E

[Alan Haisley]

Marv,
This is what I came up with:
Hard as this was, it was also fun
Alan

[Maryak]

Sine me on Cos I'm infinitely variable even to the point of cycling backwards. Others just think I'm minus one. My Dad threatened to nail my other foot to the floor to stop me going round in circles.

Best Regards
Bob

[mklotz]

Marv,
This is what I came up with:
Hard as this was, it was also fun
Alan

Congratulations, Alan, you got the correct answer.  Your method was ingenious but way too complicated.

The secret to getting an answer simply is to realize that the angle (theta in the attached diagram) made with the string must be equal on both sides.  [If they were not there would be a component of gravity operating on the weight to pull it to the position where the angles are equal.]

Armed with that insight, one can compute the angle and from there on it's just some simple trig grinding to get the answer.

[Alan Haisley]

I suspected that I had found one of the hardest ways to solve it.    But, not only did my earlier comments pretty much force me to post some kind of answer, I was sure that until I did you wouldn't post yours 
Alan

[mklotz]

I suspected that I had found one of the hardest ways to solve it.    But, not only did my earlier comments pretty much force me to post some kind of answer, I was sure that until I did you wouldn't post yours 
Alan

Hey, Alan, be proud of yourself.  Not only did you solve it and produce a verification of my equal angle assertion in the process, but you were the only one to attempt a solution.  That fact flatters not only your mathematical abilities but your character as well.

[Tennessee Whiskey]

You guys make me realize that the three years I spent in the 10th grade were all wasted. Well, maybe not, the cheerleaders were fun.

Eric

[Lew Hartswick]

For the string problem, I got 3 equations in 3 unknowns but havent gotten around to solving them yet.
   ...lew...

[Mosey]

You guys make me realize that the three years I spent in the 10th grade were all wasted. Well, maybe not, the cheerleaders were fun.

Eric

Weren't all of your years in school wasted?
Your Yankee pal,
Mosey

[mklotz]

Want to really wind your math neurons tight as a harp string?  If so, try this problem but, I warn you, this is not for the faint of heart.

A 6 unit cubical box is pushed tight against a vertical wall.  A 25 unit long ladder is placed against the wall such that it just touches the edge of the box.  What is the height of the point at which the ladder touches the wall?

Remember, the numerical answer is not the object here.  Any fool with a CAD program can find that.  What is required is the mathematical reasoning that leads to the answer.  The numerical answer is then just a quick way to check the validity of the reasoning.

[Mosey]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

[mklotz]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

Each edge is 6 units long.

[Tennessee Whiskey]

Mosey, none of them were. It's where I learned about sex, chemistry, and Rock-n-Roll. Hey, it was the 70's. At my senior prom I had the band announce that I would be buying drinks for all faculty members in the lounge. My bar bill was over $200 bucks, and that was in '77. I took an English teacher from another high school to that one. My, how times have changed. She would be on the news at 6:00 for that now.

Whiskey

[Captain Jerry]

Mosey, none of them were. It's where I learned about sex, chemistry, and Rock-n-Roll. Hey, it was the 70's. At my senior prom I had the band announce that I would be buying drinks for all faculty members in the lounge. My bar bill was over $200 bucks, and that was in '77. I took an English teacher from another high school to that one. My, how times have changed. She would be on the news at 6:00 for that now.

Whiskey

I don't see why she would be in trouble.  You were 21 then, weren't you?

Jerry

[Raggle]

Marv, I first came across that problem with the exact figures when I first got on the 'net in '98 (20ft ladder, 6ft cube) and have been entertained by it ever since, though I gained a bad reputation for asking all and sundry for an answer.

My bro-in-law, freshly in receipt of his pure maths degree, came back with a 6 line method in 2 days. But wildly wrong answer. He hadn't done the easy bit, checking with Pythagoras. He swore his maths was right.

I posed the question on a site which was set up to organise our centenary school reunion and a young whippersnapper from 2 years below me came back with a superb method but got his answer one foot out. Fell on the simple arithmetic.

No, I'm still not competent enough to do it, though I can follow this chap's method and am still in awe. I have seen a shorter method since.

I will post the whippersnapper's method, on request, but not until this thread runs its course.

Ray

[Tennessee Whiskey]

Zing, good one Jerry. Nope really just 18

[Raggle]

Oops! Your ladder is 25, mine is 20

Still, the problem is no easier in either case

Ray

[mklotz]

Ray,

Yeah, it's been around for quite a while.  An oldie but goodie, so to speak.

I'm hoping our resident genii here on the forum will come up with a spectacularly concise new solution.

When I first attempted it, I came up with a quartic equation.  It was the first time in my whole career that I had legitimately arrived at a quartic.  With some clever substitutions, though, one can work around the quartic and only have to solve quadratics.

[Captain Jerry]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

Each edge is 6 units long.

The way that I read the problem was that the 6 unit cubical box was a box of 6 cubical unit volume so  sides (and edges) would be equal to the cube root of 6. Aren't sides and edges the same? And  it seems that there should be two solutions but I don't have it yet.

Jerry

[Pete49]

the answer is......42......
Pete

[Raggle]

Here's where I found it all those years ago

http://www.stirlingsouth.com/richard/math.htm

Though the question in question is not trig

Ray

[Nickle]

This is a tough one. I got to the point of arriving at the quartic and am looking for another approach. Its been a while since I've had a shot at a set of simultaneous equations and I'm not sure those skills are still present. Lucky you didn't ask us to account for sag in the ladder due to gravity.

Regards

Nick

[Raggle]

That's why the crate is there, Nick

Ray

[mklotz]

And  it seems that there should be two solutions but I don't have it yet.

Very insightful, Jerry, and quite correct.  Think about it.  There are two ways to place the ladder so it's simultaneously touching the box, the wall and the ground.  In one position the ladder is close to vertical, reaching as far up the wall as possible.  In the other it's more nearly horizontal, with the ground end far from the wall.  Actually, these two cases are really the same case with the roles of the wall and ground interchanged.

Either solution is "correct" but by convention it's usually the first that is quoted.

[mklotz]

Aren't sides and edges the same?

No, they aren't.  A cube has 6 sides (Faces) and 12 Edges.  A point where Edges meet is called a Vertex and the cube has 8 of them.

Euler gave us a relation among these values for regular solids...

V - E + F = 2

8 - 12 + 6 = 2

As an example, a tetrahedron (4 equilateral triangle Faces) has 6 Edges and 4 Vertices.

4 - 6 + 4 = 2

Try it for an icosahedron (I don't have the patience.)

[Mosey]

This is a tough but interesting problem, but I admit that I am more interested in how light falls on the cube and shadows are cast across it by the ladder. I'm still working on the geometry.
Mosey

[Maryak]

I gave up because I couldn't remember how to manipulate quadratic equations and too lazy to look it up.

Best Regards
Bob

[Captain Jerry]

If the distance from the floor to the top of the ladder is 'A' then
 
(A-6)² + (6 + 36/(a-6) )² - 625 = 0

or A is a little more than 23 ¹/₂ but that is as far as I can go.  My eyes just slammed shut.

Jerry

[Alan Haisley]

Marv,
This one had me going for a while. After a long time demonstrating that a+b=b+a and similar silliness,    I came up with an answer that I liked. 
By the way, two other answers come to mind: 31 and 30.27 

[mklotz]

Well done again, Alan.  I'm going to have to start digging for more difficult problems in order to keep you challenged.

I hope this departure from engine minutiae has helped our members to understand what mathematics really is all about. 

[Arbalest]

I used to use these years ago but since the advent of cyano I've had no use for them.

http://www.quickwedge.com/original.htm

I had to assemble a nut, spring washer, plain washer and bolt combo on a piece of equipment this weekend where the nuts and washers were inside some box tube - out came the cyano again!

[mklotz]

Yeah, I've got several of those - well, similar, same principle, not same maker.

I've used them successfully a few times but most times they were more a source of frustration.

All screws should be Robertson style or SHCS.  Then all one needs is a magnet.

[dsquire]

Yeah, I've got several of those - well, similar, same principle, not same maker.

I've used them successfully a few times but most times they were more a source of frustration.

All screws should be Robertson style or SHCS.  Then all one needs is a magnet.

I'm in full agreement with you on that Marv.

Cheers

Don

[Arbalest]

Oh yes, I like Robbie drivers and screws but they are'nt very popular over here for some reason. I've found that Robbie fasteners "stick" onto the driver well enough on it's own, at least on the most popular red Robbie.

[black85vette]

I had one similar to this some time ago.  Liked it a lot.

http://toolmonger.com/2009/12/24/dealmonger-stanley-clip-n-grip-799/

Mine didn't have the tip that can be changed.

Still like the idea of the garrotte and need to add that to my collection of useful stuff.

PS thanks for all the cool math.  Way over my head.   I just deal with 1's and 0's and apply logic to them. There are 10 types of people in the world; those who know binary and those who don't.    :-)

[Arbalest]

Just use your standard screwdriver you always like to use but add a tiny drop of cyano (superglue) to hold the fastener temporarily in place - you won't use anything else once you've tried it!

[Captain Jerry]

Marv

So... was I right?

Jerry

[mklotz]

Marv

So... was I right?

Jerry

You're right that the answer is a bit more than 23.5 but if I substitute the correct answer A=23.673 into your equation...

(A-6)^2 + (6 + 36/(a-6) )^2 - 625 = 0

I get a value, -248.07, that's very far from zero.  Sadly, that indicates that the equation is flawed.

[Mosey]

I'm having trouble here. If the ladder is 25 long, and it touches the wall somewhere around 23.5 above the floor, then the base would be 8.54. That wouldn't allow a 6 unit cube under the ladder higher than about 3, not 6. Did I miss  something?
Mosey

[mklotz]

Draw a diagram and do a bit of checking, Mosey.

The ladder makes an angle of:

A = asin(23.673/25) = 71.248 deg

with the ground.

Armed with this value, calculate how far the bottom of the ladder is from the face of the box...

x = 6 / tan(A) = 2.037

Add that to the edge length of the box (6) and you get 8.037, which is:

8.037 = sqrt (25^2 - 23.673^2)

In cliche terms, it all adds up.

[Captain Jerry]

Marv

So... was I right?

Jerry

You're right that the answer is a bit more than 23.5 but if I substitute the correct answer A=23.673 into your equation...

(A-6)^2 + (6 + 36/(a-6) )^2 - 625 = 0

I get a value, -248.07, that's very far from zero.  Sadly, that indicates that the equation is flawed.

  RATS!

[Alan Haisley]

I wonder if an Admin could split off the math part of this and move it to "Chatterbox"? Perhaps its name there could be
Marv's Marvelous Mathematical Mysteries - Enter at your own risk.
While enjoyable as it's own thing, it shouldn't detract from the screw garrote discussion; pretty much my fault since I asked the question.
Alan

[Captain Jerry]

Marv

Due to the late hour and heavy eyes, I misstated my solution.  It should have been:

(A + 6)²  +  (6 + 36/A)² - 625 = 0

However, the value of A represents the height of the ladder above the crate
so the value substituted should be 17.673 which gives a value of .004.....

I think I got it right!

Jerry

[gordonW]

Re:- holding screws etc.  I just poke them thru' a bit of paper or card, old but works.
Re:- The maths , maybe I have mis-read this ,maybe I don't understand American, but I'm lost. Used to think I was pretty good maths. But keep 'em coming, good to think now and then.

[mklotz]

Marv

Due to the late hour and heavy eyes, I misstated my solution.  It should have been: ...

Proofreading is a virtue.

Your equation is a quartic.  Are you going to show us how you go about solving for A?

[Captain Jerry]

Marv

You asked for the reasoning, not the solution.

I'm going to go strap some wings on the donkey.  I don't think it will fly, but if it does, I will be able to show the solution.

Jerry

[mklotz]

Re:- The maths , maybe I have mis-read this ,maybe I don't understand American, but I'm lost. Used to think I was pretty good maths. But keep 'em coming, good to think now and then.

Just be glad that I didn't request that all the arithmetic be done with Egyptian fractions.

[mklotz]

Marv

You asked for the reasoning, not the solution.

I'm going to go strap some wings on the donkey.  I don't think it will fly, but if it does, I will be able to show the solution.

Jerry

If we convert your equation to its canonical quartic form, we have:

A^4 + 12*A^3 - 553*A^2 + 432*A + 1296 = 0

Plugging these coefficients into my 234 program, we get the following solutions...


Order of equation to solve (2,3,4) [2] ? 4

For order 4 term, input coefficient

• ? 1

For order 3 term, input coefficient

• ? 12

For order 2 term, input coefficient

• ? -553

For order 1 term, input coefficient

• ? 432

For order 0 term, input coefficient

• ? 1296

Solutions to:
 +1.0000 * x^4 +12.0000 * x^3 -553.0000 * x^2 +432.0000 * x^1 +1296.0000 = 0
are:
real:  17.672904  imaginary:  0.000000
real:  2.037017  imaginary:  0.000000
real:  -1.179138  imaginary:  0.000000
real:  -30.530782  imaginary:  0.000000


Note that the first solution (17.673) is the correct answer when added to the box height.  The second solution, when added to the box edge, is the (correct) distance of the bottom of the ladder from the wall.  [This is the other solution when the problem is rotated 90 degrees.]  The last two solutions are extraneous and can be disregarded.

Well done, Jerry.  In fact, it's your approach that I used when writing the program to solve this problem for arbitrary box and ladder sizes.

[Captain Jerry]

Marv

You might not believe this but the donkey took to the wings like a natural athlete.  After only two practice attempts, he gained enough altitude to clear the fence. Unfortunately, he did not have enough altitude to clear the tree line and wound up in the mid-level of an ancient live oak.  It took me most of the afternoon to talk him down.

On the way back to the house, I remembered my promise to walk you through the solution.  I guess I took too long playing with the donkey, because when I logged on to post the solution, I saw that you had already figured it out for yourself!  Nice going!

Jerry

[tel]

Well y'awl lost me a couple of pages ago BUT .... as far as I can see the problem has not been resolved at all. Take a look at the question again;

A 6 unit cubical box is pushed tight against a vertical wall.  A 25 unit long ladder is placed against the wall such that it just touches the edge of the box.  What is the height of the point at which the ladder touches the wall?

Nowhere is the height of the wall specified! Now if the height of the wall is greater than then length of the ladder, all well and good BUT if the wall happens to be shorter by any significant amount then H=h. No variation is possible.

[mklotz]

Thanks so much for that, Tel.  I just don't know what we would do without these penetrating insights of yours.  Well, actually, I do know but I'm keeping it a secret lest it elicit another penetrating insight.

[tel]

Just thought you should know - there was a risk that you might try to lean the ladder on the wall at a point avove the top, and someone might get seriously injured! 

[Mosey]

Tel,
Try to avoid complicating the logic with facts.
Mosey

[Maryak]

OK,

I used the cheats way

 

Sin B =	6/Z	Cos B =	Y/Z
Sin B =	X/A	Cos B =	6/A
Sin B =	(X+6)/25	Cos B =	(Y+6)/25
	Assign to B	71.2475	deg
0.9469	= 6/Z	0.3215	=Y/Z
Z =	0.1578	Y =	2.0370
	X+6 =	23.6729
	X =	17.6729
Tan B =	2.9455
B =	71.2475
	Solved by Iteration!!

Best Regards
Bob

[Mosey]

It can be solved rather quickly graphically (the Hardy Cross method), though obviously not exact.
Mosey

[mklotz]

It can be solved rather quickly graphically (the Hardy Cross method), though obviously not exact.
Mosey

And you'll learn nothing about mathematics by doing so.  Remember this from the original statement of the problem?

Remember, the numerical answer is not the object here.  Any fool with a CAD program can find that.  What is required is the mathematical reasoning that leads to the answer.  The numerical answer is then just a quick way to check the validity of the reasoning.