Wacko qusetion

[Car Nut]

Originally Posted by Danr55

The question indicates that the point A moved 1/8" from line A B C. In order to be 1/8" from the line, it must be in a perpendicular direction. If you move 1/8" in anything other than 90 degrees direction , the resultant distance from line A B C to the relocated point A will be less than 1/8".

I think more correctly the vector length of 1/8 inch moved, it does not say which direction and there are 360 degrees of movement that can be considered. The question indicates point A is moved 1/8 inch, (from its original point, not line ABC) as quoted "If I move point A 1/8" to the left. How much do I have to move point B to maintain a straight line to point C?"

There is not enough information to satisfy the question. That is why it is (as in the thread title) a wacko question. You cannot solve it with exactitude.

[Danr55]

What determines how far you have to move point A to achieve "1/8" to the left"?

[Car Nut]

Originally Posted by Danr55

What determines how far you have to move point A to achieve "1/8" to the left"?

Move it 1/8 inch in any direction left of point A's original position intersected by it's vertical axis (between 6 o'clock to 12 o'clock). That doesn't imply perpendicular to the axis, merely a left heading. I may have taken too much license by assuming I am not standing on my head but we will go with it.

[twobjshelbys]

Originally Posted by Mr Jody

The answer is 0".
It’s a trick question….sort of. The trick is in how you look at it.
You have to look at it as if line ABC goes from left to right. Originally, it’s 26” from A to B, and we’ll say X” from B to C, with point C being “in the distance” to the right. Something like this:
A_________________B_________________________C

After moving point A 1/8” to the left, they are all still in that same straight line ABC going from left to right. It's just that now it is 26 1/8” between A and B, and still X" between B and C. No point ever moved out of this line. Now it looks something like this (not to scale...and my attempt at a depiction just shifted B and C to the right, but you get the picture):
A_______________________B_________________________ C

That was the question I asked. If "to the left" is taken from the perspective that you are looking at the line ABC from perpendicular to the line then the answer is that B doesn't move because A' is along the original line and only the distance from AB to A'B changes. If you are looking straight into the line ABC the A' does move and I A' is a right angle to the line the new line of A'B and A'C need to be solved for the new triangle that is formed. You will first have to solve the A'AC triangle to get the new length of A'C. Then you will have to solve the A'AB to get the length of the new triangle A'BC Then you can solve the right angle of A'B to the A'C.

Again I assert that there are at least two possible answers (a different answer is to be had if the angle of A'A is not 180* or 90*) so there is insufficient information to answer it properly. Take the spec back to design engineering and ask them for clarification and engineered stamped drawings. Or overdrill B and let the drywallers patch over it.


I have a math degree... Physics too. I used to understand quantum mechanics but forgot.

[Bernica]

Originally Posted by Bernica

"If I have my surveyor's transit over point A, and I'm looking at Joe, who is holding the target at point B, and in between I have Patrick holding target C, which is directly in line with my sight of Joe, then I move my instrument to get back to my direct line of sight with Joe. Now, how much do I tell Joe to move over to get Patrick back in my direct line of sight?" Hey, I'm tryin'!

So, back to my previous example, let's say I move my transit 2 feet to my right and re-sight it in to Joe (B). Now I have direct line of sight to Joe, but I can't see Patrick (C) anymore and he can't move. Patrick wants to be back in the picture between Joe and I. Don't I just tell Joe to move 2 feet to his right as well? Is it that simple?

[Car Nut]

Originally Posted by Bernica

So, back to my previous example, let's say I move my transit 2 feet to my right and re-sight it in to Joe (B). Now I have direct line of sight to Joe, but I can't see Patrick (C) anymore and he can't move. Patrick wants to be back in the picture between Joe and I. Don't I just tell Joe to move 2 feet to his right as well? Is it that simple?

No, but it might matter if Patrick is taller.

[Bernica]

Ha! I knew I was screwed as soon as I pressed "send"

[Car Nut]

Originally Posted by twobjshelbys

I have a math degree... Physics too. I used to understand quantum mechanics but forgot.

Perhaps your trying too hard. When solving triangles you need to know 3 things, I don't think 3 things are given.

[patrickt]

Originally Posted by Car Nut

No, but it might matter if Patrick is taller.

Not only taller, but better looking and with a wonderful sense of humor as well.

[Bernica]

Originally Posted by patrickt

Not only taller, but better looking and with a wonderful sense of humor as well.

Then you could knock Tom Cruise out of the running to play Carroll Shelby in the upcoming movie!

[AL427SBF]

Originally Posted by patrickt

... During the off-season, he's guest lecturer at M.I.T. on this stuff ...

If that's the case M.I.T. students are getting screwed.

[Bernica]

I thought we were trying to solve this ourselves. You guys are way beyond my level of competence, but I never quit and I don't ask for help (a determent, I know). I could call in Caltech friends to fight against MIT, but that takes it out of our hands.

Maybe I should just ponder on .035 vs .038 plug gaps, proper ignition systems, tires, twin roll bars etc.

[AL427SBF]

Originally Posted by Three Peaks

If A, B, and C are in a straight line to begin with, the amount of movement of point C would equal 1/8" x the ratio of the distances between AB and BC (AB/BC=???? x 1/8")

Bob

I think you are on the right track but the C movement has to be less than 1/8" so BC/AB x 1/8.

The more convoluted approach ..

A'
.......................................C'

A---------- ----------------C---------------------------------------B

If he is at A looking down the line to B with C in between then ...
Moving to the left 1/8" equates to moving A up 1/8". Now he can see B unobstructed by C.

Q: How far do you have to move C up (C') to be back in line with the new A'B line.

The angle created by A' to B back to A is the same angle as C' to B back to A, call it theta.

Tangent (theta) = (opposite/adjacent) = (1/8")/26" = .00481
Arctan .00481 = 0.00480996291 rad = 0.275590574 degrees

Without knowing where C lies in the original AB line you can't solve for the distance to move C to C' (call it Z). If you did know the distance of CB then ...

Tan (0.275590574) = Z/CB
Z = Tan (0.275590574) x CB

[Bernica]

You can't move Patrick (C)

[AL427SBF]

Originally Posted by Bernica

You can't move Patrick (C)

Nor would I want to, not even with a 10' pole.

[Ralphy]

So am I guessing right, your not looking for a number? Your wanting the equation/formula no matter where/distance C is? Damn so rusty in this ****! Hell I couldn't answer even I weren't! Does this have anything to do with where you put your beer?

Ralphy

[twobjshelbys]

Originally Posted by Car Nut

Perhaps your trying too hard. When solving triangles you need to know 3 things, I don't think 3 things are given.

Another hitch in the saddle. It wasn't stated whether it was "up". I'm thinking 10th grade planar geometry. So the answer could be that A'' is "higher". That wouldn't change the solution I described except the answer for B' could be "higher" rather than to the East. (or west, or south or north, depending on where you're standing when you determine which angle to the line A' is...)

Again, cry foul and say need more info

[twobjshelbys]

Originally Posted by AL427SBF

I think you are on the right track but the C movement has to be less than 1/8" so BC/AB x 1/8.

The more convoluted approach ..

A'
.......................................C'

A---------- ----------------C---------------------------------------B

If he is at A looking down the line to B with C in between then ...
Moving to the left 1/8" equates to moving A up 1/8". Now he can see B unobstructed by C.

Q: How far do you have to move C up (C') to be back in line with the new A'B line.

The angle created by A' to B back to A is the same angle as C' to B back to A, call it theta.

Tangent (theta) = (opposite/adjacent) = (1/8")/26" = .00481
Arctan .00481 = 0.00480996291 rad = 0.275590574 degrees

Without knowing where C lies in the original AB line you can't solve for the distance to move C to C' (call it Z). If you did know the distance of CB then ...

Tan (0.275590574) = Z/CB
Z = Tan (0.275590574) x CB

I read that C was the fixed point and B/B' was the midpoint....


Overdrill...

[mikiec]

OK, it's time to stop the misery.

You have a straight line with points A B C along this line. Point A is at one end point C is at the other end. Point B is 26 inches from point A. Point B to point C is 10'(I think I used 10') Point A is moved to the left 1/8"

To maintain a straight line from ABC point B must move to the left. How far does point B move?

[AL427SBF]

A'
.......................................B'

A---------- ----------------B---------------------------------------C


Tangent (theta) = (opposite/adjacent) = (1/8")/146" = .000856
Arctan .000856 = 0.000855999791 rad

Tan (0.000855999791 rad) = Z/120"
Z = Tan (0.000855999791 rad) x 120"
Z = 0.000856 x 120"
Z = .103" (rounded)