# need help with algebra



## Tinstaafl (Jan 6, 2008)

neolitic said:


> I thought he asked for the equation,
> not the *answer*. :laughing:


Well, since the equation IS the answer, the answer is the answer too. :blink:


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## neolitic (Apr 20, 2006)

I thought he could just tell
Asimo what he wanted, and
the parts would pop right outta 
that machine...:laughing:


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## Gus Dering (Oct 14, 2008)

neolitic said:


> I thought he asked for the equation,
> not the *answer*. :laughing:
> 
> 
> ...


You are so correct. I'm still working on it.

Although having the answer helps in testing the endless amount of bad equations.

I have an equation that works but I can't for the life of me break it down.

This is what I got;

.707X + X = 48 

This works if you plug in the correct answer of 28.125.

does any one know how to get to 28.125 from that equation?


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## Warren (Feb 19, 2005)

I think that has to be expressed in feet and inches. Let me ask my 13 year old.


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## Tinstaafl (Jan 6, 2008)

Gus Dering said:


> does any one know how to get to 28.125 from that equation?


Wuss. Do the quadratic. :laughing:

.707X + X = 48

1.707X = 48

Divide both sides by 1.707

X = 28.11950790861159929701230228471


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## Warren (Feb 19, 2005)

I still was the one that came up with the answer first! Not bad for a guy who never took math past the tenth grade!:clap:


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## mics_54 (Oct 28, 2008)

eveyone knows the equation A² + B²= c²


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## Tinstaafl (Jan 6, 2008)

mics_54 said:


> eveyone knows the equation A² + B²= c²


Yeah... That was the beginning logic for that quadratic solution.

What are YOU doing with it?


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## Gus Dering (Oct 14, 2008)

Warren said:


> I still was the one that came up with the answer first! Not bad for a guy who never took math past the tenth grade!:clap:


Well thats cool. They teached this stuff in 9th grade 30 sumpin years ago.



Tinstaafl said:


> Wuss. Do the quadratic. :laughing:
> 
> .707X + X = 48
> 
> ...


That is what I'm talking about. Lone's equation is the net result as well but can you explain in plain terms the logic behind the move from .707X + X = 48 to 1.707X = 48?

I got there by way of geometric relationship of a right triangle with equal legs.

The Pythagorean therom is in play but when the two legs of a right triangle are equal the diagonal is = to the leg x the square root of 2. The ol, 1 , 1 sq root of 2. same as 3 , 4 , 5. A sq + B sq = C sq , same thing


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## loneframer (Feb 13, 2009)

Gus Dering said:


> You are so correct. I'm still working on it.
> 
> Although having the answer helps in testing the endless amount of bad equations.
> 
> ...


 I thought that's what I did, no?:blink:


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## Tinstaafl (Jan 6, 2008)

.707X + X is the same thing as X plus part (.707) of another X. That equals 1X plus .707X, which equals 1.707X.

And .707 is the sine of 45°.


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## Tinstaafl (Jan 6, 2008)

loneframer said:


> I thought that's what I did, no?:blink:


You did, but you ASSumed everyone would read between the lines. :thumbsup:

Frankly, it's been too long since I did any real trig, so I'm faking my way through this with rudimentary algebra. :jester:


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## katoman (Apr 26, 2009)

Warren said:


> I still was the one that came up with the answer first! Not bad for a guy who never took math past the tenth grade!:clap:


Congratulations, you get the orio cookie :thumbup:


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## loneframer (Feb 13, 2009)

Tinstaafl said:


> You did, but you ASSumed everyone would read between the lines. :thumbsup:
> 
> Frankly, it's been too long since I did any real trig, so I'm faking my way through this with rudimentary algebra. :jester:


 I never learned either formally, that's how I ended up in construction.:whistling What math I do know, I learned on the jobsite, so maybe it's not textbook in form, but it is elementary in function.:laughing:


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## Gus Dering (Oct 14, 2008)

Tinstaafl said:


> .707X + X is the same thing as X plus part (.707) of another X. That equals 1X plus .707X, which equals 1.707X.
> 
> *And .707 is the sine of 45°*.


It's also 1/2 the square root of 2. Which is how I got there before I even started this thread. I just couldn't make the jump to the 1.707 part.

Thanks for your explanation. 

Now I have to get ready for the pep rally and find a senior to buy my beer. :laughing:


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## Tinstaafl (Jan 6, 2008)

I never really "understood" trig until I got a job in a machine shop and applied it in the real world. :thumbsup:

Now that i think about it, is Gus maybe asking how/why the sine of the angle comes into play here? I'm on the fuzzy edge of that, but not close enough to explain it.


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## Gus Dering (Oct 14, 2008)

loneframer said:


> I thought that's what I did, no?:blink:


You were the closest to the real answer out of this mottly crew. I'll give you that. 

I was stumped in that breakdown is all. You are and always will be the man among men.:notworthy


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## Gus Dering (Oct 14, 2008)

Tinstaafl said:


> I never really "understood" trig until I got a job in a machine shop and applied it in the real world. :thumbsup:
> 
> Now that i think about it, is Gus maybe asking how/why the sine of the angle comes into play here? I'm on the fuzzy edge of that, but not close enough to explain it.


I go draw another picture. I'll be back


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## Tinstaafl (Jan 6, 2008)

Gus Dering said:


> You were the closest to the real answer out of this mottly crew. I'll give you that.


I beg to differ. <sniff>

My frickin' savant son had the best solution because it's such a headache-maker to understand. :laughing:

Seriously, both solutions are quite accurate--but Lone's is WAY easier to actually work out. :notworthy


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## loneframer (Feb 13, 2009)

Here's my attempt at explaining the .707107, easier done with a 1" unit. The diagonal of 1" at a 45 is 1.414214". That needs to be converted back to the linear dimension rather than the 45, so divide the unit of 1" by its diagonal of 1.414214". For every inch of straight run, you need .707107" of straight run to achieve 1" on the diagonal.

In Gus' case, 61" - 13" of cabinet depth leaves 48". You need to devide 48" by a combination of 1" units for the straight run and an equal amount of .707107" units for the angled run.

48" devided by 1.707107" = 28.11775"

48" - 28.11775" =19.88225" the diagonal of which is 28.11775"

Again, I'm no seasoned mathemagician, but it gets the job done.:thumbsup:


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## Warren (Feb 19, 2005)

I can usually figure out a problem like that faster than I can explain how. I used a variation of Pythagorean theory as well. I looked at it as a roof problem. We know that the hip run is always 16.97 vs the 12 of a common. Basically, I took an educated guess, based on my observation of the diagram, and adjusted until I got the correct answer.


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## Tinstaafl (Jan 6, 2008)

loneframer said:


> Here's my attempt at explaining the .707107, easier done with a 1" unit. The diagonal of 1" at a 45 is 1.414214". That needs to be converted back to the linear dimension rather than the 45, so divide the unit of 1" by its diagonal of 1.414214". For every inch of straight run, you need .707107" of straight run to achieve 1" on the diagonal.


That is an *excellent* explanation! This is the sort of thing they tried to teach [some of us] back in high school, but only by means of rote memorization and trig tables. If only they'd have presented us with real-world applications like this, the light would have clicked on way sooner.

You da man! :clap:


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## Tinstaafl (Jan 6, 2008)

Warren said:


> I can usually figure out a problem like that faster than I can explain how.


That's the case for most of us. OTOH, when i taught some basic electronics courses 100 years ago, I found that explaining things helped to solidify and organize everything in my own mind, and I was subsequently much more efficient at solving those sorts of problems.

In a way, it's the "student teaching the teacher", and everyone benefits. We should all have apprentices. :thumbsup:


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## loneframer (Feb 13, 2009)

Tinstaafl said:


> That is an *excellent* explanation! This is the sort of thing they tried to teach [some of us] back in high school, but only by means of rote memorization and trig tables. If only they'd have presented us with real-world applications like this, the light would have clicked on way sooner.
> 
> You da man! :clap:


 Thanks bro. I sucked at math in HS. I just didn't get how it applied to anything. Fortunately, as a framer, the applications started to click and I was fortunate enough to be working for a guy who was able to make me understand a little of the math. 

Like I said, I wasn't formally taught, so my ways of getting there may only work for me.:thumbsup:


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## Gus Dering (Oct 14, 2008)

View attachment triangle.bmp


I may not give this explanation justice but I'll give it a whirl.

With equal legs and using A sq + B sq = C sq we have that triangle with a diagonal of 1.414 .

So then the leg is equal to .707x.

.707X + X = 48

I know I am missing something in there but it makes good sence in my head.:laughing:


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## Tinstaafl (Jan 6, 2008)

loneframer said:


> Like I said, I wasn't formally taught, so my ways of getting there may only work for me.:thumbsup:


Obviously not. Look at Gus's post right below yours. :laughing::thumbsup:



Gus Dering said:


> I know I am missing something in there but it makes good sence in my head.:laughing:


Gus, you have it. That drawing is a perfect complement to both your and Lone's explanations. :thumbsup:

My idiot son just came home, saw Lone's solution and said "Duh! I wish they'd have given us problems like this in school!" :laughing:


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## Gus Dering (Oct 14, 2008)

I got a chance to do a career day at a high school for kids that were on their last chance so to speak. I loved it, wish I could do more. The biggest message I had to say was that I use my high school math every day at my job and I am a better professional for it.

I'm not sure why more schools don't invite people to the class to explain how these theories pertain to real world challenges.


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## Tinstaafl (Jan 6, 2008)

Gus Dering said:


> I'm not sure why more schools don't invite people to the class to explain how these theories pertain to real world challenges.


Sadly, I suspect it's because the staff themselves don't have much of a clue. Most of them went to school, then went to school to learn how to teach in school. :blink:


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## Gus Dering (Oct 14, 2008)

I think I will reach out and see if I can find a geometry teacher that is receptive to an old nail bender like myself. I think that would be fun.:thumbsup:


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## loneframer (Feb 13, 2009)

loneframer said:


> Here's my attempt at explaining the .707107, easier done with a 1" unit. The diagonal of 1" at a 45 is 1.414214". That needs to be converted back to the linear dimension rather than the 45, so divide the unit of 1" by its diagonal of 1.414214". For every inch of straight run, you need .707107" of straight run to achieve 1" on the diagonal.
> 
> In Gus' case, 61" - 13" of cabinet depth leaves 48". You need to devide 48" by a combination of 1" units for the straight run and an equal amount of .707107" units for the angled run.
> 
> ...


This approach would work well if the ends of the units were angled as well.

Suppose you wanted 2 outside corners, an inside corner and a straight run with 2 modules. All you need to do is multiply .707107 by three for the angled segments and add 2 for the straight modules. 4.121321

61 devided by 4.121321=14.80108 which is the width of each straight module. So two straight modules =29.602", which leaves 31.39784" for three angled modules., or 10.46595" per module, the diagonal of which is 14.80108. This would allow for all doors to be exactly the same for all modules, assuming they are not full overlay, causing the inside corner door to bind.:whistling


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## Tinstaafl (Jan 6, 2008)

Now that you found it, you just have to keep playing with it, don't you?

Keep it up and you'll go blind. :laughing:


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## Gus Dering (Oct 14, 2008)

loneframer said:


> This approach would work well if the ends of the units were angled as well.
> 
> Suppose you wanted 2 outside corners, an inside corner and a straight run with 2 modules. All you need to do is multiply .707107 by three for the angled segments and add 2 for the straight modules. 4.121321
> 
> 61 devided by 4.121321=14.80108 which is the width of each straight module. So two straight modules =29.602", which leaves 31.39784" for three angled modules., or 10.46595" per module, the diagonal of which is 14.80108. This would allow for all doors to be exactly the same for all modules, assuming they are not full overlay, causing the inside corner door to bind.:whistling


Now if you could just come to the front and draw that on the chalk board for us please.:thumbsup: 



Tinstaafl said:


> Now that you found it, you just have to keep playing with it, don't you?
> 
> Keep it up and you'll go blind. :laughing:


He does spend an insane amount of time in the shower.:laughing:


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## neolitic (Apr 20, 2006)

Now if someone would explain why
FF goes to a new screen every time
I use "ALT" to enter an ASCII code.....


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## loneframer (Feb 13, 2009)

Tinstaafl said:


> Now that you found it, you just have to keep playing with it, don't you?
> 
> Keep it up and you'll go blind. :laughing:


Now you sound like my mother.:whistling


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## Tinstaafl (Jan 6, 2008)

neolitic said:


> Now if someone would explain why
> FF goes to a new screen every time
> I use "ALT" to enter an ASCII code.....


Never got around to trying FF, but have you checked Preferences for hotkey combos? You might be triggering one of them.


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## loneframer (Feb 13, 2009)

Gus Dering said:


> Now if you could just come to the front and draw that on the chalk board for us please.:thumbsup:
> 
> 
> 
> He does spend an insane amount of time in the shower.:laughing:


 You don't really want me to start drawing the things I see in my mind, do you?:shutup:

Keeping it clean is a full time proposition.:laughing:


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## Gus Dering (Oct 14, 2008)

neolitic said:


> Now if someone would explain why
> FF goes to a new screen every time
> I use "ALT" to enter an ASCII code.....


I wish I knew how to do what is that you are trying to do. If I knew I might know what to say so then you would know, sadly I do not know.:no:



loneframer said:


> Now you sound like my mother.:whistling


 If you knew how to post a picture, we would all have a better understanding of what you know.

Oh yeah, you do know..


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## neolitic (Apr 20, 2006)

Tinstaafl said:


> Never got around to trying FF, but *have you checked Preferences for hotkey combos?* You might be triggering one of them.


I would if there was such an option.
If there is one, it is well hidden.


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## Tinstaafl (Jan 6, 2008)

neolitic said:


> I would if there was such an option.
> If there is one, it is well hidden.


Yeah, Google reveals that it doesn't exist. Nor should that keycombo be hitting any of the ones that are built in. Cluemeter = zero. :sad:


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## loneframer (Feb 13, 2009)

Tinstaafl said:


> Now that you found it, you just have to keep playing with it, don't you?
> 
> Keep it up and you'll go blind. :laughing:


 Suppose you want to build an octagonal structure, 16' wide. What is the segment length of each side?

192" devided by 2.414214 or(( 2*.707107)+1)= 79.52899"

192-79.52899= 112.571 devided by 2= 56.23551, which the diagonal of is 79.52902. (Within 1/1000 is fairly close.)

Works great for laying out footings, floor joists, rafters...


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