How to Rearrange formula? Solve 2(Pc - d) / Pc for Pc

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
Hello,

Having trouble rearranging formula to solve for Pc. Formula is 2(Pc - d) / Pc.

Any ideas would be appreciated.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,134
Hello,

Having trouble rearranging formula to solve for Pc. Formula is 2(Pc - d) / Pc.

Any ideas would be appreciated.
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
How to rearrange formula

I know the basic rules for rearrangement, but don't quite understand when its not set against and = sign, and I have Pc in the dividend and divisor.

1) 2(Pc - d) / Pc

2) 2Pc - 2d / Pc

3) Pc * (2Pc - 2d) / Pc * Pc

4) Pc(2Pc - 2d)

5) 2Pc Squared - 2dPc

It just doesn't make any sense to me.....
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
2(Pc - d) / Pc
I'm assuming that Pc is a single symbol (not P times c).

This is not an equation, so there is nothing to solve. This is an expression. We can rewrite expressions (sometimes called simplifying), but I wouldn't necessarily say that the following version is simpler.

2(1 - d/Pc)

Please post the entire exercise statement, including all instructions, word for word.
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
I'm assuming that Pc is a single symbol (not P times c).

This is not an equation, so there is nothing to solve. This is an expression. We can rewrite expressions (sometimes called simplifying), but I wouldn't necessarily say that the following version is simpler.

2(1 - d/Pc)

Please post the entire exercise statement, including all instructions, word for word.
This is a formula for Boiler inspectors to determine allowable circumferential pitch of boiler tubes.
d = 3.03125" (Boiler tube diameter)
Pc = ??? (Circumferential pitch of adjacent openings in ")

Problem: The drum of a bent tube type boiler is drilled for tube holes 3.03125" diameter spaced longitudinally in groups of two tubes pitched 4.875 with 6" spaced between groups. Circumferentially tube holes are symetrical. What is the circumferential pitch of the tubes. Solve for Pc
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
I googled some keywords, and I found a number of sites from which you may download ASME Sec. 1 (2010).

The longitudinal pitch between the staybolts, or between the nearest row of staybolts and the row of rivets at the joints between the furnace sheet and the tube sheet or the furnace sheet and mud ring, shall not exceed that given by the following formula: 56320t2L=2PR where L = longitudinal pitch of staybolts, P = maximum allowable working pressure (pounds per square inch), t = thickness of furnace sheet (inches), R = outside radius of furnace (inches). When values by this formula are less than the circumferential pitch, the longitudinal pitch may be as large as the allowable circumferential pitch. The stress in the staybolts shall not exceed 7500 psi. and shall be determined as specified in PFT-23.4.23.4 In furnaces over 38 in. in outside diameter and combustion chambers not covered by special rules in this Section which have curved sheets subject to pressure on the convex side, neither the circumferential nor longitudinal pitches of the staybolts shall exceed 1.05 times that given by the rules in PG-46.
That is an example of a formula. :cool:
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
I googled some keywords, and I found a number of sites from which you may download ASME Sec. 1 (2010).

That is an example of a formula. :cool:
This formula is directly out of the ASME Section 1 (2017) book. Is it possible to solve for Pc?
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
A = 2(Pc-d)/Pc
This equation can be solved for Pc.

I'll get you started.

1) Multiply each side by Pc

2) Use the Distributive Property, to multiply out 2(Pc - d)
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
This formula is directly out of the ASME Section 1 (2017) book. Is it possible to solve for Pc?
Yes. Are you familiar with how to solve one-variable linear equations? Thank you! ;)
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
This equation can be solved for Pc.

I'll get you started.

1) Multiply each side by Pc

2) Use the Distributive Property, to multiply out 2(Pc - d)
From my earlier post: I don't quite understand when I have Pc in the dividend and divisor.

1) 2(Pc - d) / Pc

2) 2Pc - 2d / Pc

3) Pc * (2Pc - 2d) / Pc * Pc

4) Pc(2Pc - 2d)

5) 2Pc Squared - 2dPc

It just doesn't make any sense to me..... I am starting to think it is impossible to actually solve for Pc with supplied info.
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
If you had followed my suggestion, you would have:

A = 2(Pc - d) / Pc

A·Pc = 2(Pc - d)

A·Pc = 2·Pc - 2·d

Now, we want to get all terms containing Pc to one side of the equation, with all terms that do not contain Pc on the other side. There's more than one way to do this, but the end result will be the same.

I choose to add 2·d to each side, and then subtract A·Pc from each side.

2·d = 2·Pc - A·Pc

Now we can factor out Pc, on the right-hand side.

Can you finish it?
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
If you had followed my suggestion, you would have:

A = 2(Pc - d) / Pc

A·Pc = 2(Pc - d)

A·Pc = 2·Pc - 2·d

Now, we want to get all terms containing Pc to one side of the equation, with all terms that do not contain Pc on the other side. There's more than one way to do this, but the end result will be the same.

I choose to add 2·d to each side, and then subtract A·Pc from each side.

2·d = 2·Pc - A·Pc

Now we can factor out Pc, on the right-hand side.

Can you finish it?
Don't think I can, but..: When you say factor out Pc, could I drop the Pc that comes after the A making the 2Pc a 3Pc? such as: 2*d = 3*Pc - A?
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
If you had followed my suggestion, you would have:

A = 2(Pc - d) / Pc

A·Pc = 2(Pc - d)

A·Pc = 2·Pc - 2·d

Now, we want to get all terms containing Pc to one side of the equation, with all terms that do not contain Pc on the other side. There's more than one way to do this, but the end result will be the same.

I choose to add 2·d to each side, and then subtract A·Pc from each side.

2·d = 2·Pc - A·Pc

Now we can factor out Pc, on the right-hand side.

Can you finish it?
I think what would make more sense to me than previous post is 2*d = 2*Pc - A*Pc, Factor out Pc from both terms such as 2*d = Pc - A. Is this correct?
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
When you say factor out Pc, could I drop the Pc that comes after the A making the 2Pc a 3Pc?
Nope. Factoring is the reverse of distribution.

Distributive Property:

A(B - C) = A·B - A·C

Instead, if we start with the expression A·B - A·C, then we can factor out the A, and get back to where we started.

Factoring:

A·B - A·C = A(B - C)
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
I think what would make more sense to me than previous post is 2*d = 2*Pc - A*Pc, Factor out Pc from both terms such as 2*d = Pc - A. Is this correct?
2·d = 2·Pc - A·Pc

On the right-hand side, factor out Pc:

2·d = Pc (2 - A)

Last step: divide each side by (2 - A)

Check out this site, for some basic lessons. Or, google keywords solve linear equations.
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
Nope. Factoring is the reverse of distribution.

Distributive Property:

A(B - C) = A·B - A·C

Instead, if we start with the expression A·B - A·C, then we can factor out the A, and get back to where we started.

Factoring:

A·B - A·C = A(B - C)

So 2*d = 2*Pc - A*Pc would change to 2*d = Pc(2-A). Then divide both sides by 2-A. Answer equals Pc = 2d/2-A?
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,251
So 2*d = 2*Pc - A*Pc would change to 2*d = Pc(2-A). Then divide both sides by 2-A.

Answer equals Pc = 2d/(2-A)?
That's correct. Here's a note about typing math.

When we write the algebraic ratio \(\displaystyle \frac{2d}{2 - A}\) with paper and pencil, it's easy to see the denominator of the ratio because we use a horizontal fraction bar.

We cannot type a horizontal fraction bar, so we must use grouping symbols (shown above in red) to make clear what's on the top (numerator) and what's on the bottom (denominator).

Typing the expression 2d/2-A means this:

\(\displaystyle \dfrac{2d}{2} - A\)

But typing 2d/(2 - A) means this:

\(\displaystyle \dfrac{2d}{2 - A}\)

If you're a student or you need to do algebra at work, I recommend that you take the time to study beginning algebra. There are lots of free courses and textbooks available on-line. Cheers, and watch out for those hot bolts. 8-)
 

SeabeeDavid

New member
Joined
Jan 17, 2018
Messages
11
That's correct. Here's a note about typing math.

When we write the algebraic ratio \(\displaystyle \frac{2d}{2 - A}\) with paper and pencil, it's easy to see the denominator of the ratio because we use a horizontal fraction bar.

We cannot type a horizontal fraction bar, so we must use grouping symbols (shown above in red) to make clear what's on the top (numerator) and what's on the bottom (denominator).

Typing the expression 2d/2-A means this:

\(\displaystyle \dfrac{2d}{2} - A\)

But typing 2d/(2 - A) means this:

\(\displaystyle \dfrac{2d}{2 - A}\)

If you're a student or you need to do algebra at work, I recommend that you take the time to study beginning algebra. There are lots of free courses and textbooks available on-line. Cheers, and watch out for those hot bolts. 8-)
MMM4444BOT, Thank you for your assistance, I am a Boiler Inspector Trainee getting ready to take certification exam in the next month. There are various formulas for figuring a lot of information out within the ASME for construction of boilers, but that one is the one I was having problems with. I have taken the algebra courses, but don't use very often. Thank you very much and I appreciate your time.
 
Top