134a Refrigerant
#4081
Guest
Posts: n/a
Re: 134a Refrigerant
"Nathan W. Collier" <MontanaJeeper@aol.com> wrote in message
news:11esia5htubde55@corp.supernews.com...
> "Stephen Cowell" <scowell@sbcglobal.net> wrote in message
> news:eFpHe.1139$5q7.489@newssvr19.news.prodigy.com ...
> >> exactly. i grossed over a grand today alone. what did you do, loser?
> > :-)
> >
> > Do you mean what did I "do", or what did I "earn"?
>
> both, as im quite sure that neither will amount to much above zero.
> > And which day, exactly, are you referring to?
>
> .....dumb *** cant read a date on a message header.
You made $1000 gross on Sunday? Well, I certainly
must admit that I didn't make a dime on Sunday.
Must put me into pretty hard company, for sure.
__
Steve
..
#4082
Guest
Posts: n/a
Re: 134a Refrigerant
OK here's the equation for calculating the sticking volume of a
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
#4083
Guest
Posts: n/a
Re: 134a Refrigerant
OK here's the equation for calculating the sticking volume of a
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
#4084
Guest
Posts: n/a
Re: 134a Refrigerant
OK here's the equation for calculating the sticking volume of a
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
#4085
Guest
Posts: n/a
Re: 134a Refrigerant
OK here's the equation for calculating the sticking volume of a
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
storage tank.
xx
x x
x------------x Liquid level which is variable,
x x can go up or down
x x
x x
x x
xx
Imagine the above is a circle. I know the diameter of the circle,
and I can measure from the top of the circle down to the liquid level.
How can I use this information to derive the amount of liquid in the
circle? I hope to make a formula I can use in a ------sheet.
Thanks so much.
--------------------------------------------------------------------------------
Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?
Hello there!
Thanks for writing to Dr. Math! You asked a good question.
Before we start, let's go over the "law of sines" (we'll be using
it shortly). Given a triangle ABC where A, B, and C are vertices,
the following relation holds:
sin A sin B sin C
-------- = ------- = -------
A B C
where sin A means sin of the angle at vertex A, and A (when it
appears in the denominator) is the length of the side of the
triangle opposite A.
Okay, now on to the problem. You did a nice job with your drawing
on the computer, so I'm not going to draw another picture. Instead,
you draw a picture on paper and label the following:
Call the distance from the top of the circle to the liquid level h.
Call the radius of the circle r.
(these are given quantities)
Now, draw in the center of the circle. Draw a line connecting the
center of the circle to the top of the circle. Then draw two more
lines from the center of the circle to the points where the liquid
level lines intersect the circle. Note that these lines are just radii,
so they have length r.
Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y. Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.
By the law of sines, we have the following relations:
r r-h x
----- = -------- = -----
* sin 90 sin y sin z
sin 90 = 1, so we have:
r-h
----- = r
sin y
Solve this equation for y to get:
y = arcsin [(r-h)/r]
Note also that z = 90 - y.
And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] =
(1/r)Sqrt(2rh - h^2)
(this is by trig identities and stuff...if you have questions,
write back).
So, then equation * becomes:
rx
r = --------------
Sqrt (2rh - h^2)
So, x = Sqrt (2rh - h^2)
Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment
we've been talking about that is of angle 2z from the center.
The area of this segment occupied by liquid is just the area of
the 2 small triangles. We know the base of each triangle is x
(see above) and the height is r-h. So, the total area contributed
by this segment is x(r-h).
Now consider the area outside of the segment. That is, we want
the area of a portion of the circle. If we measure z in degrees we
want the portion of the circle that takes up 360 - 2z degrees of the
circle. So, the fraction we are dealing with here is:
360 - 2z
--------
360
The area of the portion we want then is:
360 - 2z
-------- (Pi)r^2
360
Now add these two areas together and you are done. Thus, the total
area is going to be:
360 - 2(90 - arcsin [(r-h)/r])
------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2)
360
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/
Stephen Cowell wrote:
>
> You continue to top yourself! Are you not aware
> of the formula for a regular cylinder? Do you not
> agree that an area times a length gives you your
> cubic dimension? What about good old Mr.
> What'sHisName? Didn't he teach you about this?
> You claimed to have 'gone through Calculus'!
> Like corn through a goose, I replied... apropos.
>
> Here's another hint... the liquid level forms
> a chord across the end, parallel to a diameter
> of the circular face. Draw a line from the
> center of the circular face to the edge of the
> tank where the liquid level is... a radius line.
> It should all be clear now... if you can figure
> out the area of a piece of 2D pie!
> __
> Steve
> .
#4086
Guest
Posts: n/a
Re: 134a Refrigerant
"Stephen Cowell" <scowell@sbcglobal.net> wrote in message
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
#4087
Guest
Posts: n/a
Re: 134a Refrigerant
"Stephen Cowell" <scowell@sbcglobal.net> wrote in message
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
#4088
Guest
Posts: n/a
Re: 134a Refrigerant
"Stephen Cowell" <scowell@sbcglobal.net> wrote in message
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
#4089
Guest
Posts: n/a
Re: 134a Refrigerant
"Stephen Cowell" <scowell@sbcglobal.net> wrote in message
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
news:j0XHe.547$646.168@newssvr22.news.prodigy.net. ..
> We all read it... you gave an example
> of 'in a pinch' as 'such as on a rooftop in -30 degree weather'...
you stupid stool. were it not for low ambient the HMC wouldnt be needed AT
ALL. its funny to watch you squirm and spin, hoping desperately that you
can convince use that you actually know something about refrigeration. LOL!
:-)
> No... you gave an example of 'in a pinch'. This in no
> way defined the exact terms of the challenge
lol --------, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)
> why do you attemp to put such a condition onto me,
> when you didn't make the same condition to Jeff?
being stump-stupid, you require further clarification. :-)
> Says a lot about you, doesn't it? Says 'welch' to me..
you pathetic little liar! i outlined it for you BEFORE you even knew what
an HMC was.
>> oh yeah, a link with "proudliberal" in the URL is credible. <rolling
> eyes>
>
> Logical fallacies
lol.....youre actually trying to suggest that a URL including "proudliberal"
is credible? BUWHAHAHA! :-)
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
#4090
Guest
Posts: n/a
Re: 134a Refrigerant
"Stephen Cowell" <scowell@sbcglobal.net> wrote in message
news:W5XHe.549$646.45@newssvr22.news.prodigy.net.. .
> You can't give up
lol, not a chance! :-)
> We all can read... and remember.
your lies are there for one and all.
>> LIAR. i asked youf or credible sources and you give me sites with
>> "proudliberal" in the url. i know why you would stretch for such a lie
> but
>> nobody is falling for it.
>
> Now you're very confused again
TRANSLATION --> "its lying time"
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com
news:W5XHe.549$646.45@newssvr22.news.prodigy.net.. .
> You can't give up
lol, not a chance! :-)
> We all can read... and remember.
your lies are there for one and all.
>> LIAR. i asked youf or credible sources and you give me sites with
>> "proudliberal" in the url. i know why you would stretch for such a lie
> but
>> nobody is falling for it.
>
> Now you're very confused again
TRANSLATION --> "its lying time"
--
Nathan W. Collier
http://InlineDiesel.com
http://7SlotGrille.com
http://UtilityOffRoad.com
http://BighornRefrigeration.com