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Originally posted by ac rookie
I think I am going into another field. I will never figure this stuff out!!!

It's not hard. Actually, to accomplish what Mid wants to accomplish, you spend fifty bucks on a ductulator that does this FOR you...

Originally posted by Steve Wiggins
I guarantee you this will seperate the Dems from the Repubs right quick. We will always have the thinkers vs. the doers.

Are you lost again, buttcheek?

Originally posted by Andy Schoen

Originally posted by glennwith2ns

Originally posted by mattm
You big bunch of math nerds.

Yea , bet they got pocket protectors too>)

And only the best pocket protectors... We're still trying to figure out where to put that Parker name on them...

Do me a favor-------------DON'T put it on there.

Originally posted by glennwith2ns

Originally posted by mattm
You big bunch of math nerds.

Yea , bet they got pocket protectors too>)

And only the best pocket protectors... We're still trying to figure out where to put that Parker name on them...

Originally posted by midhvac
But I wanted to know how to find the diameter when CFM and pd are known. With your help, I've got it now. My $8.99 Casio scientific calculator won't allow me to use a decimal point in a fractional exponent, so I couldn't do ^1/5.02. So I just changed it from a fraction to a decimal equivalent of .1992

dia = (.109136 x CFM^1.9/pd)^.1992

You've got the idea.

If I'm reading your equation correctly (I really should look it up), you can get precise answers doing the following on your Casio:

calculate A:

A = CFM^1.9

next B:

B = .109136 x A /pd

next C:

C = LN(B) this is the natural log of B (I'm assuming the Casio has log functions)

next D:

D = C / 5.02

finally E:

E = exp(D) this is the e^x function which is inverse to the natural log

I guarantee you this will seperate the Dems from the Repubs right quick. We will always have the thinkers vs. the doers.

Thanks again Andy. I got it now. The reason for my questions was that I was trying to figure out how the different ductulator formulas work so I could use my scientific calculator to do them, or plug them into an Excel spreadsheet application. Somebody else here was asking about the formulas a while back, but nobody knew them. I found a formula that calculates the pressure drop per 100' when the cfm and duct diameter are known:

pd = (.109136 x CFM^1.9)/dia^5.02

But I wanted to know how to find the diameter when CFM and pd are known. With your help, I've got it now. My $8.99 Casio scientific calculator won't allow me to use a decimal point in a fractional exponent, so I couldn't do ^1/5.02. So I just changed it from a fraction to a decimal equivalent of .1992

dia = (.109136 x CFM^1.9/pd)^.1992

Originally posted by mattm
You big bunch of math nerds.

Yea , bet they got pocket protectors too>)

You big bunch of math nerds.

I think I am going into another field. I will never figure this stuff out!!!

Originally posted by midhvac
Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?

Nope. Using apropriate math:

a = (2 x 20^4) / b^6

a = 320,000 / b^6

b^6 = 320,000 / a

(b^6)^(1/6) = (320,000 / a)^(1/6)

b = (320,000 / a)^(1/6)

Originally posted by midhvac

Originally posted by condenseddave
Mid-

You are going to sprain your brain if you keep this up.

Nah, I put an elastic support around my head, just in case.

Let's put that elastic support to work...

Using your calculator, solve the following:

x^3 = x * x * x = -27

or x = (-27)^(1/3)

A little figuring will show that 'x' must equal -3, i.e., -3 * -3 * -3 = -27

Why does the calculator fail to figure this out?

Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?

Originally posted by condenseddave
Mid-

You are going to sprain your brain if you keep this up.

Nah, I put an elastic support around my head, just in case.

You guys rock! Thanks! I've got a scientific calculator and the 1/6 power works.

Originally posted by midhvac
In this formula b = 6

a = (2 x 20^4) / b^6
So the answer is 6.8587

Perhaps some number theory may be appropriate here.

'a' in this example is equal to exactly 320,000 / 46,656 or approximately 6.85871056241426611796982167352538 (using Microsoft calculator )

'a' is considered a 'rational' number because it can be wriiten as a fraction, i.e., x / y where x and y are integers.

Originally posted by midhvac

If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?

Ok, lets assume 'a' = 7, what is 'b'?

b = (320,000 / 7)^(1/6)

where 'b' is approximately 5.979643919487405903197485402773 (again using Microsoft calculator)

In this case, 'b' is an irrational number. There is no fraction, x / y, that can define 'b' where x and y are integers. Think about it! No integer numbers whatsoever. Raising this expression to the 1/6th power causes this to occur.

Mmmmm... so does this help?

midhvac,

Do the basic algebra and up end up with:

b=(2x20^4/a)all raised to the 1/6

You need a scientific calculator to get what b is. That is, 46656.07 ^1/6

plug in all the numbers and the answer is 6.

Is this what you were asking?

Mid-

You are going to sprain your brain if you keep this up.

Holy cow...your asking me to remember my pre calculus days. When your trying to solve to that power things get tuff. Well, I will tell you right now Im not going to do it!





































Im glad thats over.

In this formula b = 6

a = (2 x 20^4) / b^6
So the answer is 6.8587

If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?

If it did come out as something to the something power, how would I convert that to a regular number?