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Main
Date: 24 May 2006 06:46:29
From: Ryan Case
Subject: points per pound per gallon?
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Someone mentioned this the other day. I haven't had time to properly
research it on the web yet, so was hoping that someone here would be
kind enough to enlighten me on it.
Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
100% efficient extraction? Giving me a wort of 1.074? And then my real
hydrometer reading / 1.074 would give me my efficiency, right? So if I
actually read 1.065 after the boil, my efficiency would be 87%?
(Obviously dropping the 1's and decimal to make the math nicer)
Thanks for any help. I realize that this is actually a pretty simple
concept, but something tells me I am missing something.
Ryan
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Date: 24 May 2006 09:20:23
From: Scott L
Subject: Re: points per pound per gallon?
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Ryan Case wrote:
> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
> 100% efficient extraction? Giving me a wort of 1.074? And then my real
> hydrometer reading / 1.074 would give me my efficiency, right? So if I
> actually read 1.065 after the boil, my efficiency would be 87%?
> (Obviously dropping the 1's and decimal to make the math nicer)
Your gravity calculation is correct. Your number for the efficiency is
also correct, but you aren't just dropping the 1 from the readings to
make the math nicer. If you don't drop the 1, you get a completely
bogus number. I think of it in terms of points and just take 65 / 74.
If you get confused using ppg, write everything out using the units and
make sure the units cancel to get the one you want. Then you know
you've automatically got it right.
Scott
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Date: 24 May 2006 10:50:15
From: Dan Listermann
Subject: Re: points per pound per gallon?
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"Ryan Case" <usenet@jamesrobert.us > wrote in message
news:1278otlpb8dosd7@corp.supernews.com...
> Someone mentioned this the other day. I haven't had time to properly
> research it on the web yet, so was hoping that someone here would be kind
> enough to enlighten me on it.
>
> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
> 100% efficient extraction? Giving me a wort of 1.074? And then my real
> hydrometer reading / 1.074 would give me my efficiency, right? So if I
> actually read 1.065 after the boil, my efficiency would be 87%? (Obviously
> dropping the 1's and decimal to make the math nicer)
>
> Thanks for any help. I realize that this is actually a pretty simple
> concept, but something tells me I am missing something.
>
> Ryan
It looks like you have the basic idea.
Dan
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Date: 24 May 2006 07:57:34
From: Ryan Case
Subject: Re: points per pound per gallon?
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Dan Listermann wrote:
> "Ryan Case" <usenet@jamesrobert.us> wrote in message
> news:1278otlpb8dosd7@corp.supernews.com...
>> Someone mentioned this the other day. I haven't had time to properly
>> research it on the web yet, so was hoping that someone here would be kind
>> enough to enlighten me on it.
>>
>> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
>> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
>> 100% efficient extraction? Giving me a wort of 1.074? And then my real
>> hydrometer reading / 1.074 would give me my efficiency, right? So if I
>> actually read 1.065 after the boil, my efficiency would be 87%? (Obviously
>> dropping the 1's and decimal to make the math nicer)
>>
>> Thanks for any help. I realize that this is actually a pretty simple
>> concept, but something tells me I am missing something.
>>
>> Ryan
>
> It looks like you have the basic idea.
>
> Dan
>
>
OK then. Now for a harder one. Hop Utilization. How does one figure the
amount of IBU passed on to the beer?
I know that it is a function of the IBU or AA% rating of the hop and the
amount of time it is in the boil, but what is that function?
Ryan
Who just feels more at ease if he understands these things. It's not
that I won't use a program like promash, but I feel like I am wandering
around aimlessly if I don't at least grasp the math that the program is
doing. Make sense?
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Date: 24 May 2006 11:45:38
From: Dan Listermann
Subject: Re: points per pound per gallon?
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"Ryan Case" <usenet@jamesrobert.us > wrote in message
news:1278t2v71joduec@corp.supernews.com...
> OK then. Now for a harder one. Hop Utilization. How does one figure the
> amount of IBU passed on to the beer?
>
> I know that it is a function of the IBU or AA% rating of the hop and the
> amount of time it is in the boil, but what is that function?
There are a lot of different formulas that hours have been spent debating
over. They can become very complex. Frankly, almost no homebrewers
actually measure their bitterness so who is to say which is better? Here is
the one I use. It is simple enough that many times I can work it in my
head.
IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
For five gallons, the constant is just 4.
The constant assumes an hour boil, more than an hour can be ignored, less
should be scaled.
It is surprising how well this works out compared to fancier formulas. The
difference would rarely cause a significant change in a recipe.
Dan
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Date: 24 May 2006 09:38:19
From: Denny Conn
Subject: Re: points per pound per gallon?
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Dan Listermann wrote:
> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
Dan, I have to thank you for that formula! I've been using it (and
teaching it in BJCP study groups) since you first posted it years ago!
--------- >Denny
--
Life begins at 60 - 1.060, that is.
Reply to denny_at_projectoneaudio_dot_com
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Date: 24 May 2006 13:45:45
From: Dan Listermann
Subject: Re: points per pound per gallon?
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Thanks for saying that! When you work at a homebrew shop, it is a handy
thing to use. Usually I can do it in my head, but sometimes I want a
calculator.
In my earlier example of 35 IBU from 6%AA hops in five gallons, I would
think 35 / 4 = almost 9 , 9 / 6 = 1.5 oz
Dan
"Denny Conn" <me@privacy.net > wrote in message
news:44748BFB.D8500D07@privacy.net...
> Dan Listermann wrote:
>
>> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
>
> Dan, I have to thank you for that formula! I've been using it (and
> teaching it in BJCP study groups) since you first posted it years ago!
>
> --------->Denny
>
> --
> Life begins at 60 - 1.060, that is.
>
> Reply to denny_at_projectoneaudio_dot_com
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Date: 24 May 2006 09:18:24
From: Ryan Case
Subject: Re: points per pound per gallon?
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Dan Listermann wrote:
> "Ryan Case" <usenet@jamesrobert.us> wrote in message
> news:1278t2v71joduec@corp.supernews.com...
>> OK then. Now for a harder one. Hop Utilization. How does one figure the
>> amount of IBU passed on to the beer?
>>
>> I know that it is a function of the IBU or AA% rating of the hop and the
>> amount of time it is in the boil, but what is that function?
>
> There are a lot of different formulas that hours have been spent debating
> over. They can become very complex. Frankly, almost no homebrewers
> actually measure their bitterness so who is to say which is better? Here is
> the one I use. It is simple enough that many times I can work it in my
> head.
>
> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
>
> For five gallons, the constant is just 4.
>
> The constant assumes an hour boil, more than an hour can be ignored, less
> should be scaled.
>
> It is surprising how well this works out compared to fancier formulas. The
> difference would rarely cause a significant change in a recipe.
>
> Dan
>
>
I'm sorry, and maybe I am being dense, but where does the constant go in
the equation? Are you just dividing by four at the end to assume a 25%
absorption rate?
Thanks for the replies, I am working on getting to a point where I can
reproduce some of my better beers and would like to understand some of
this stuff a little better for that purpose.
Ryan
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Date: 24 May 2006 13:41:38
From: Dan Listermann
Subject: Re: points per pound per gallon?
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"Ryan Case" <usenet@jamesrobert.us > wrote in message
news:12791qgrda89b13@corp.supernews.com...
> Dan Listermann wrote:
>> "Ryan Case" <usenet@jamesrobert.us> wrote in message
>> news:1278t2v71joduec@corp.supernews.com...
>>> OK then. Now for a harder one. Hop Utilization. How does one figure the
>>> amount of IBU passed on to the beer?
>>>
>>> I know that it is a function of the IBU or AA% rating of the hop and the
>>> amount of time it is in the boil, but what is that function?
>>
>> There are a lot of different formulas that hours have been spent debating
>> over. They can become very complex. Frankly, almost no homebrewers
>> actually measure their bitterness so who is to say which is better? Here
>> is the one I use. It is simple enough that many times I can work it in
>> my head.
>>
>> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
>>
>> For five gallons, the constant is just 4.
>>
>> The constant assumes an hour boil, more than an hour can be ignored, less
>> should be scaled.
>>
>> It is surprising how well this works out compared to fancier formulas.
>> The difference would rarely cause a significant change in a recipe.
>>
>> Dan
> I'm sorry, and maybe I am being dense, but where does the constant go in
> the equation? Are you just dividing by four at the end to assume a 25%
> absorption rate?
20 is the constant. 20 / 5 = 4.
Say you want to make five gallons of 35 IBU beer using 6% AA hops. 35 = 6
* oz * 20 / 5 Solving for oz, you get 1.45 oz of hops. I call it 1.5 and
be done with it.
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Date: 24 May 2006 11:03:38
From: Ryan Case
Subject: Re: points per pound per gallon?
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Dan Listermann wrote:
> "Ryan Case" <usenet@jamesrobert.us> wrote in message
> news:12791qgrda89b13@corp.supernews.com...
>> Dan Listermann wrote:
>>> "Ryan Case" <usenet@jamesrobert.us> wrote in message
>>> news:1278t2v71joduec@corp.supernews.com...
>>>> OK then. Now for a harder one. Hop Utilization. How does one figure the
>>>> amount of IBU passed on to the beer?
>>>>
>>>> I know that it is a function of the IBU or AA% rating of the hop and the
>>>> amount of time it is in the boil, but what is that function?
>>> There are a lot of different formulas that hours have been spent debating
>>> over. They can become very complex. Frankly, almost no homebrewers
>>> actually measure their bitterness so who is to say which is better? Here
>>> is the one I use. It is simple enough that many times I can work it in
>>> my head.
>>>
>>> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
>>>
>>> For five gallons, the constant is just 4.
>>>
>>> The constant assumes an hour boil, more than an hour can be ignored, less
>>> should be scaled.
>>>
>>> It is surprising how well this works out compared to fancier formulas.
>>> The difference would rarely cause a significant change in a recipe.
>>>
>>> Dan
>> I'm sorry, and maybe I am being dense, but where does the constant go in
>> the equation? Are you just dividing by four at the end to assume a 25%
>> absorption rate?
>
> 20 is the constant. 20 / 5 = 4.
>
> Say you want to make five gallons of 35 IBU beer using 6% AA hops. 35 = 6
> * oz * 20 / 5 Solving for oz, you get 1.45 oz of hops. I call it 1.5 and
> be done with it.
>
>
You see, I was being dense. So since I almost never bother with anything
other then 10 gallons now, I could replace everything after "oz *" with
"2" and simplify it even more, no?
Oh, and for scaling it for boil time. I just multiply my final IBU
number for the hops by (boil time/60) right?
Thanks again.
Ryan
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Date: 24 May 2006 14:39:04
From: Dan Listermann
Subject: Re: points per pound per gallon?
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"Ryan Case" <usenet@jamesrobert.us > wrote in message
news:12797vr7fib25f7@corp.supernews.com...
> Dan Listermann wrote:
>> "Ryan Case" <usenet@jamesrobert.us> wrote in message
>> news:12791qgrda89b13@corp.supernews.com...
>>> Dan Listermann wrote:
>>>> "Ryan Case" <usenet@jamesrobert.us> wrote in message
>>>> news:1278t2v71joduec@corp.supernews.com...
>>>>> OK then. Now for a harder one. Hop Utilization. How does one figure
>>>>> the amount of IBU passed on to the beer?
>>>>>
>>>>> I know that it is a function of the IBU or AA% rating of the hop and
>>>>> the amount of time it is in the boil, but what is that function?
>>>> There are a lot of different formulas that hours have been spent
>>>> debating over. They can become very complex. Frankly, almost no
>>>> homebrewers actually measure their bitterness so who is to say which is
>>>> better? Here is the one I use. It is simple enough that many times I
>>>> can work it in my head.
>>>>
>>>> IBU = AA * oz * 20 / gal. (AA is alpha acid rating of the hops)
>>>>
>>>> For five gallons, the constant is just 4.
>>>>
>>>> The constant assumes an hour boil, more than an hour can be ignored,
>>>> less should be scaled.
>>>>
>>>> It is surprising how well this works out compared to fancier formulas.
>>>> The difference would rarely cause a significant change in a recipe.
>>>>
>>>> Dan
>>> I'm sorry, and maybe I am being dense, but where does the constant go in
>>> the equation? Are you just dividing by four at the end to assume a 25%
>>> absorption rate?
>>
>> 20 is the constant. 20 / 5 = 4.
>>
>> Say you want to make five gallons of 35 IBU beer using 6% AA hops. 35 =
>> 6 * oz * 20 / 5 Solving for oz, you get 1.45 oz of hops. I call it 1.5
>> and be done with it.
>>
>>
>
> You see, I was being dense. So since I almost never bother with anything
> other then 10 gallons now, I could replace everything after "oz *" with
> "2" and simplify it even more, no?
2 would be correct for ten gallons.
>
> Oh, and for scaling it for boil time. I just multiply my final IBU number
> for the hops by (boil time/60) right?
You got it!
Dan
>
> Thanks again.
>
> Ryan
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Date: 25 May 2006 14:41:08
From: John 'Shaggy' Kolesar
Subject: Re: points per pound per gallon?
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On Wed, 24 May 2006 07:57:34 -0700, <usenet@jamesrobert.us > wrote:
> OK then. Now for a harder one. Hop Utilization. How does one figure the
> amount of IBU passed on to the beer?
>
> I know that it is a function of the IBU or AA% rating of the hop and the
> amount of time it is in the boil, but what is that function?
>
> Ryan
>
> Who just feels more at ease if he understands these things. It's not
> that I won't use a program like promash, but I feel like I am wandering
> around aimlessly if I don't at least grasp the math that the program is
> doing. Make sense?
Hop utilization is more complicated. In addition to AA% and time you also
have things like the SG during the boil which are a factor. There are
basically three formulas for calculating IBU, all of which tend to get
a little complex. IMO, let promash do this for you.
John.
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Date: 24 May 2006 13:36:16
From: E
Subject: Re: points per pound per gallon?
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"Dan Listermann" <dan@listermann.com > wrote in message
news:1278sl3iiubhf6e@corp.supernews.com...
>
> "Ryan Case" <usenet@jamesrobert.us> wrote in message
> news:1278otlpb8dosd7@corp.supernews.com...
>> Someone mentioned this the other day. I haven't had time to properly
>> research it on the web yet, so was hoping that someone here would be kind
>> enough to enlighten me on it.
>>
>> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
>> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
>> 100% efficient extraction? Giving me a wort of 1.074? And then my real
>> hydrometer reading / 1.074 would give me my efficiency, right? So if I
>> actually read 1.065 after the boil, my efficiency would be 87%?
>> (Obviously dropping the 1's and decimal to make the math nicer)
>>
>> Thanks for any help. I realize that this is actually a pretty simple
>> concept, but something tells me I am missing something.
>>
>> Ryan
>
> It looks like you have the basic idea.
May I ask a stupid question. What does the 37 represent and where do you
get it from? I understand the rest but I'm new to trying to understand the
homebrew thing and I'm not sure how you go about finding out the 37 number
(I assume it depends on what grains / how much grain is in the recipe?).
Thank you.
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Date: 24 May 2006 10:53:45
From: Ryan Case
Subject: Re: points per pound per gallon?
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E wrote:
> "Dan Listermann" <dan@listermann.com> wrote in message
> news:1278sl3iiubhf6e@corp.supernews.com...
>> "Ryan Case" <usenet@jamesrobert.us> wrote in message
>> news:1278otlpb8dosd7@corp.supernews.com...
>>> Someone mentioned this the other day. I haven't had time to properly
>>> research it on the web yet, so was hoping that someone here would be kind
>>> enough to enlighten me on it.
>>>
>>> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
>>> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
>>> 100% efficient extraction? Giving me a wort of 1.074? And then my real
>>> hydrometer reading / 1.074 would give me my efficiency, right? So if I
>>> actually read 1.065 after the boil, my efficiency would be 87%?
>>> (Obviously dropping the 1's and decimal to make the math nicer)
>>>
>>> Thanks for any help. I realize that this is actually a pretty simple
>>> concept, but something tells me I am missing something.
>>>
>>> Ryan
>> It looks like you have the basic idea.
>
> May I ask a stupid question. What does the 37 represent and where do you
> get it from? I understand the rest but I'm new to trying to understand the
> homebrew thing and I'm not sure how you go about finding out the 37 number
> (I assume it depends on what grains / how much grain is in the recipe?).
>
> Thank you.
>
>
37 is the laboratory points per gallon that I used in this example for
the imaginary grain. It also happens to be the points per gallon of most
pale 2-row. This is of course if you were 100% efficient.
Pale 2-row malt 37
Pale 6-row malt 35
Munich malt (Belgian/German) 37
Munich malt (US) 33
Wheat malt 39
Crystal malt 34
Special "B" 30
Chocolate Malt 30
Black Patent 29
Roasted Barley 29
Amber/Brown malt 32
Cara-pils (US) 33
Cara-pils (Belgian) 34
Aromatic 36
Biscuit 35
Ryan
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Date: 24 May 2006 13:56:21
From: E
Subject: Re: points per pound per gallon?
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"Ryan Case" <usenet@jamesrobert.us > wrote in message
news:12797da58n1hpea@corp.supernews.com...
Ah, thanks. Now I understand.
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Date: 25 May 2006 14:38:05
From: John 'Shaggy' Kolesar
Subject: Re: points per pound per gallon?
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On Wed, 24 May 2006 06:46:29 -0700, <usenet@jamesrobert.us > wrote:
> Someone mentioned this the other day. I haven't had time to properly
> research it on the web yet, so was hoping that someone here would be
> kind enough to enlighten me on it.
>
> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
> 100% efficient extraction? Giving me a wort of 1.074? And then my real
> hydrometer reading / 1.074 would give me my efficiency, right? So if I
> actually read 1.065 after the boil, my efficiency would be 87%?
> (Obviously dropping the 1's and decimal to make the math nicer)
>
> Thanks for any help. I realize that this is actually a pretty simple
> concept, but something tells me I am missing something.
You've pretty much got it right. Keep in mind though that minor errors in
measuring the weight of grain used and the volume in the kettle can really
throw off your efficiency numbers. Don't get too caught up in what your
efficiency number actually is. It's much more important to have a consistent
efficiency than it is to have a really high one. I would much prefer to
have a 70% on every batch than to have ones that range between 80 - 90 all
the time.
John.
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Date: 25 May 2006 07:56:05
From: Ryan Case
Subject: Re: points per pound per gallon?
|
John 'Shaggy' Kolesar wrote:
> On Wed, 24 May 2006 06:46:29 -0700, <usenet@jamesrobert.us> wrote:
>> Someone mentioned this the other day. I haven't had time to properly
>> research it on the web yet, so was hoping that someone here would be
>> kind enough to enlighten me on it.
>>
>> Say I have 20lbs of grain in my bill for a 10 gallon batch. The grain is
>> rated at 37. Does that mean that 20 * 37 / 10 should be my perfect world
>> 100% efficient extraction? Giving me a wort of 1.074? And then my real
>> hydrometer reading / 1.074 would give me my efficiency, right? So if I
>> actually read 1.065 after the boil, my efficiency would be 87%?
>> (Obviously dropping the 1's and decimal to make the math nicer)
>>
>> Thanks for any help. I realize that this is actually a pretty simple
>> concept, but something tells me I am missing something.
>
> You've pretty much got it right. Keep in mind though that minor errors in
> measuring the weight of grain used and the volume in the kettle can really
> throw off your efficiency numbers. Don't get too caught up in what your
> efficiency number actually is. It's much more important to have a consistent
> efficiency than it is to have a really high one. I would much prefer to
> have a 70% on every batch than to have ones that range between 80 - 90 all
> the time.
>
>
> John.
I am not too caught up in it trust me. In fact I have been home brewing
on and off for the better part of ten years and have only owned a
hydrometer for the last two months.
I am just interested in learning a bit more about the math that is
involved in it so that I can go at producing consistent results armed
with a little knowledge.
Also, if I can get my efficiency nailed down to a consistent number like
you were saying then I can better work out recipes.
Ryan
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Date: 25 May 2006 15:42:50
From: John 'Shaggy' Kolesar
Subject: Re: points per pound per gallon?
|
On Thu, 25 May 2006 07:56:05 -0700, <usenet@jamesrobert.us > wrote:
> I am not too caught up in it trust me. In fact I have been home brewing
> on and off for the better part of ten years and have only owned a
> hydrometer for the last two months.
>
> I am just interested in learning a bit more about the math that is
> involved in it so that I can go at producing consistent results armed
> with a little knowledge.
>
> Also, if I can get my efficiency nailed down to a consistent number like
> you were saying then I can better work out recipes.
No problem, sounds like you're approaching it the right way. It's just that
some brewers seem to get caught up in a bragging contest over who's got the
highest efficiency, which is kind of silly. You're right, a consistent
efficiency will help you to fine tune your recipes.
John.
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