Losing Weight: a theory
Note:
A calorie is the heat energy required to raise the temperature of 1 kilogram of water by 1 degree centigrade.

I've been thinking about losing weight and ...

>Just thinking?
Pay attention.
I found the following Harris-Benedict Equation for BMR (Basal Metabolic Rate):
 Women: BMR = 655 + ( 4.35 x weight in pounds ) + ( 4.7 x height in inches ) - ( 4.7 x age in years ) Men: BMR = 66 + ( 6.23 x weight in pounds ) + ( 12.7 x height in inches ) - ( 6.8 x age in years )
or, if you prefer:
 Women: BMR = 655 + ( 9.6 x weight in kg ) + ( 1.8 x height in cm ) - ( 4.7 x age in years ) Men: BMR = 66 + (13.7 x weight in kg ) + ( 5 x height in cm ) - ( 6.8 x age in years )
Note that they each have the form: a + b (Weight) + c (Height) - d (Age)

Multiply by 1.2 and you get the number of calories you burn if you're sleeping
... no excercise, no calorie-burning activity, just "at rest".
Multiply by 1.375 and you get the calories needed if you're lightly active.
Multiply by 1.9 and you get the calories needed if you're very active.
We can call the number 1.2 or 1.375 or 1.9 the exercise factor.

>The number of calories burned while sleeping?
Yes ... in order to maintain your current body weight. Your body needs energy just to survive, eh?
Eat less and you lose weight. Eat more and ...

>Ya, I get it it, but do you actually believe that number?
Irrelevant. I just want to play with the magic formula.
Note that there are 3500 calories in a pound of stored body fat. We're going to need that number.
It's a fictitious number, to be sure, but everybody uses it ... incorrectly! So will we.

 Current Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) calories per day "at rest" calorie consumption
Calculator #1
1.2 x Basal Metabolic Rate
Okay, here we go:
• Suppose you eat K calories per day.
• According to the above BMR formula, you burn f (a + bW + c (height) - d (age) ) calories per day
where W is your weight and the other numbers, a, b, c and d are constants and f is your exercise factor.
• The net amount is then K - f (a + bW + c (height) - d (age) ) calories per day.
• If this is positive, it generates body fat. If it's negative, you lose body fat.
>At the rate of 1 pound of body fat for every 3500 calories, right?
You got it!
So, dividing by 3500 gives the rate at which your weight changes:
 [1]     dW/dt = [ K - f (a + bW + c (height) - d (age) ) ] / 3500   pounds per day.
This differential equation has the form:
 [2]     dW/dt = A - B W   with solution   W(t) = A/B + (W0 - A/B) e-B t
Solutions looks like Figure 1

>Aah ... there's some eventual weight, right?
Yes ... and it's this guy:
 [3]     Eventual "Equilibrium" Weight = A/B = [ K - f (a + c (height) - d (age) ] / ( f b )

Note: Each of A and B have a factor 3500 ... but it cancels out when taking the ratio A/B.
Further, your current weight doesn't enter the result, either!

Figure 1

You eat too much or exercise too little
Okay, what we're really interest in is: What's the best strategy: Eat less or exercise more?

>Or both!
Yes ... so we consider the "eventual weight" (denoted above by A/B) which we'll call the Equilibrium Weight.
It's the weight consistent with your calorie intake.
Decrease that intake and your weight eventually drops to another Equilibrium Weight.

>Increase it and you gain weight.
Yes, eventually ... until you reach some larger Equilibrium Weight (as depicted in Figure 1).

So play with the calculator and tell me whether you should eat less or exercise more.
Which gives a greater decrease in weight? Which is the most "effective"? Which provides ...

>I think that eating less loses weight faster ... but you ain't as healthy.
Think so? Personally, I like the "at rest" option, with beer and pretzels in front of the TV.

 Current Weight pounds (optional) Height ft   in Age Male or Female ? (0 for Male, 1 for Female) Intake calories pounds "at rest" equilibrium weight
Calculator #2
Equilibrium Weight doesn't depend upon your Current Weight.
Okay, let's do this:
• Assume we're male, weigh 200 pounds, 5' 10"" in height, 60 years old ... and we're lethargic.
• Stick these numbers into the Calculator #1 and get the "at rest" Calorie Consumption of 2152.
That's what we'd consume in order to maintain our 200 pound weight - without exercise.
• Stick these numbers into Calculator #2 to confirm that our Equilibrium Weight is, indeed, 200 pounds.
• Now reduce the Calorie Intake in Calculator #2 from 2152 to 2052 and get a new "at rest" equilibrium weight of 187 pounds.
A daily 100 calorie reduction will (eventually) reduce our 200 pound weight to 187 pounds.
• Did you check out what a "lightly active" lifestyle would have given?
In fact, it'll get you a new equilibrium weight of 163 pounds.

>Aha! So do some exercise a few times a week, engage in some sports, jog and ...
... and end the day with beer and pretzels in front of the TV.

>And you believe all this mathematical manipulation?
Uh ... well, no ... but its fun.
In fact, I purposely neglected to put any time scale in the chart of Figure 1. If I did, we'd see that it'd take several years to reach that equilibrium weight.
That's way too long ... and it doesn't even consider the fact that losing 3500 calories means losing 1 pound. (Actually, the 3500 number is was too optimistic anyway!)

If you stare at magic equation [1] you see the 3500 value. For a sedentary 200 pound, 60 year old guy with 5' 10" height munching 2152 calories each day, the right side of [1] is:
( K - f (a + bW + c (height) - d (age) ) ) / 3500   or   ( 2152 - 1.2*(66 + 5.23*200 + 12.7*70 - 6.8*60 ) ) / 3500 = 0.0687 pounds per day.
Now (as suggested above), he reduces his daily calorie-munching by 100 calories... to 2052 calories per day.
 He ends up with a weight of 187 pounds (as noted above). Do you have any idea how long that would take? >I'd say, as a rough guess ... let's see, he lost 13 pounds at 0.0687 pounds per day so that'd be ... uh ... 189 days. Well the weight loss depends upon the weight, which is continually decreasing -- so it ain't that easy. However, from Figure 2 you can see that he's down to 188 pounds in over 1000 days. Mathematically speaking, it's take forever to reach that eventual equilibrium weight. It's better to see how long it takes to get close to the Equiilibrium Weight. Figure 2 Weight loss over time
Check out equation [2]. If we call the Equilibrium Weight EW (so EW = A/B), then it says:
 [2a]     W / EW - 1 = (W0/EW - 1) e-B t   or, taking logs (to the base e), we have: [2b]     T = log[ (W0/EW-1) / (W/EW-1) ] / B = the time to go from weight Wo to weight W
If we use the numbers from our example, namely Wo = 200 pounds and W= 188 pounds, we'd get (since B = 0.002136 in our example):

T = log[ (200/187-1) / (188/187-1) ] / 0.002136 = 1200 days, the time to go from weight Wo to weight W.

>Where'd that B-value come from?
You mean B = 0.002136? Look very carefully at equations [1] and [2] and you'll see that:
 [4] A = ( K - f (a + c (height) - d (age) ) ) / 3500 B = f b / 3500
where K = Intake Calories and f depends upon your activity,
running from f = 1.2 to f = 1.9
and the other parameters depend upon various biological factors,
namely:
 Parameter Male Female a 66 655 b 6.23 4.35 c 12.7 4.7 d 6.8 4.7
There's a calculator here
I wouldn't place much faith in the numbers generated ... but they're interesting, no?

>NO !

I might point out that your weight will always be between your Current Weight and the Equilibrium Weight.
See Figure 1.

If you ask for a Desired Weight outside that range, then ... uh ...

 Current Weight pounds (optional) Desired Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) Intake calories days to reach desired weight using "at rest" equilibrium weight
Calculator #3
T = Time to reach desired weight.

 Losing Weight: a (simple) theory

Okay, let's proceed differently.

1. We'll use the above prescription for the number of calories you burn to maintain your current weight, namely Calculator #1. (Example: 2152 calories.)
2. Then we ask how much weight you'd like to lose. (Example: 10 pounds.)
3. Then we ask how long you wish to wait until you achieve that weight loss. (Example: 70 days)
4. Then we use the (fallacious) "fact" that a reduction of 3500 calories will lose just 1 pound. (Example: For 10 pounds, you need to lose 35,000 calories ... in 70 days.)
5. Then we'll calculate much you must reduce your daily calorie intake to achieve that loss. (Example: 35,000 / 70 = 500 calories per day.)

>"Fact"?
Let's take 3500 / pound as "fact" ... even if it ain't. Can't do much without numbers, eh?
Note that step 1 isn't necessary.
It's just there to remind you what you now burn (to maintain your Current Weight) and what the Desired Weight Loss will require.

>I like it! It's simple!
Yes, but I lost 10 pounds in 10 days.
According to this "simple" ritual, that'd be a reduction of 3500 calories per day.
Fat chance!

I prefer the earlier mathematical gesticulation.
The only thing that bothers me is the time it takes to lose, say, 10 pounds.
For my situaltion, it'd take months, not days.

However, note that (in [3], above), the parameter called B scales the time variable.
If we increase B, losing weight occurs at a faster rate.
Alas, it also changes the Equilibrium Weight, namely A / B.

 Current Weight pounds Desired Weight Loss pounds Number of Days days Height ft   in Age Male or Female ? (0 for Male, 1 for Female) calories reduction per day Assuming "sedentary" life style
Calculator #4
Daily calorie reduction to achieve your desired weight loss.
 However, we can fiddle with that -- to retain the Equlibrium Weight (and everything else), but speed things up. >You think mathematical manipulation will lose wieght faster? I don't know what you're drinking, but ... No, I'm trying to get a mathematical prescription that'll be in line with my losing 10 pounds in 10 days ... without reducing my calorie intake by 3500 calories each day. Take a peek at Figure 3. Note that the eventual Equilibrium Weight don't hardly change ... but we get there fatser, eh? >Fatser? Is that Freudian slip? We get there faster. >Ya, but can you justify that increased rate? I'm working in it ... but do you see that faint grey curve, in Figure 3? That's the result of assuming you'd lose a pound for every 2500 calorie reduction ... instead of 3500. Figure 3 Changing the time scale
 >So you think the number 3500 calories per pound of fat is wrong?? No, that's not what I'm saying. We can't argue with that number. You stick food into a bomb calorimeter, press a button and get the heat energy (or "calories, which derives from the Latin word calor = heat). Bomb calorimeter
 In fact, there's this thing called MEI (Metabolic Energy Intake). You take 85% of the calories contained in a food item and assume that's what the body extracts from the food. Indeed, it's the result of multiplying the actual calories (as per calorimeter) by 0.85 that generates the labels on food. They're MEI-adjusted calories and more accurately reflect what the body takes from the food item. In particular, that 3500 becomes, after MEI adjustment, 2975 calories. Let's say 3000. The result is to increase the rate at which you lose weight. (See that faint grey curve, in Figure 3?) Note that the Equlibrium Weight does't change. That depends upon A / B and the number of calories required to lose a pound cancels out in the ratio. MEI-adjusted calories on food labels
If we change that 3500 to 3000, then Calculator #3 gives these results:

Comparing with the earlier result, the time required to lose 10 pounds goes from 318 days to 273 days.

>Yeah ... but what about 10 pounds in 10 days?
Uh ... yes, that change (3500 to 3000) won't give me the my 10-day time period to lose 10 pounds.

 Current Weight pounds (optional) Desired Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) Intake calories days to reach desired weight using "at rest" equilibrium weight
Calculator #3A (using 3000 calories per pound)
T = Time to reach desired weight.

 Losing Weight: Metabolic Efficiency

I figure that two guys with identical parameters (age, height and weight) can have dramatically different weight losses -- even though they reduce their calorie intake by the same amount. One guy uses more of the calories consumed to generate body fat. He's more "efficient".

It's like that Metabolic Energy Index, where only 85% of the calorie intake is used to generate a pound of body fat. That 85% is some kind of "efficiency".

Suppose we search for a "Calories per pound of weight loss" that'll come close to 10 pounds in 10 days.
We can try this calculator. Just stick in some number (like X = 2500) to see what you get:

>And you believe all this stuff?
I take it all with a very large grain of salt.

>Salt? I understand the maximum recommended level of sodium intake is 2,300 mg per day, so ...
You're not paying attention.

 Current Weight pounds (optional) Desired Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) Intake calories X = Calories per pound calories days to reach desired weight using "at rest" equilibrium weight
Calculator #3C (using X calories per pound)
T = Time to reach desired weight.

>So what's this Metabolic Efficiency stuff?
Well ... I figure the an "efficient" guy uses a greater percentage of the calories consumed to lose or gain a pound. He's got a higher efficiency rate.
You increase your intake by 3000 calories in a week and use it all ... my body may use only 2500 of that 3000 calories to generate a pound.
I have a smaller Metabolic Efficiency.

So let's incorporate that into our calculator, eh?

 Current Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) calories per day "at rest" calorie consumption
Check your "at rest" Calorie Consumption

 Current Weight pounds (optional) Desired Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) Intake calories ME = Metabolic Efficiency % days to reach desired weight using "at rest" equilibrium weight
Calculator #3D (using ME: Metabolic Efficiency)
T = Time to reach desired weight.

See how it works? You reduce the ME and, although your Equlibrium Weight don't hardly change, you lose weight faster.
In fact, that ME acts just like the MEI. We assume that:
• A change in intake of 3000 calories is needed to gain (or lose) a pound: that's the MEI-reduced value.
• A very ME-efficient person will use nearly all of them 3000 calories for each pound.
• A person with, say, ME = 50% will use just 50% of the 3000 calories to gain (or lose) a pound.
Note:
Beware the 9999999 result. Your "Desired Weight" must be between your "Current Weight" and your "Equlibrium Weight".
You get a 9999999 and it means you'll never reach the Equilibrium Weight.
Microscopic!
In fact, if I stick in ME = 3.5%, I can lose 10 pounds in 10 days.
If I stick in ME = 0%, I can lose 10 pounds in a microsecond!

>Wait a minute! If the "normal", 3000 MEI-adjusted calorie intake gets you a pound, then 3.5% of that is only 105 calories. Are you saying that ...
I'm saying that, from 3000 (MEI-adjusted) calories per pound of body dat, I need only 105 calories to change my weight by a pound.
>And you call that "Efficiency"? Sounds more like inefficiency to me.
Uh ... maybe so.

Note, however, that we've been using that Harris-Benedict Equation which was created in 1919.
There are newer, perhaps more relevant formulas. Note, however, that BMR depends upon lots of factors:

• People with a large percentage of muscle will have a larger Basal Metaboic Rate. They burn lots of calories but it has a lesser effect on their body fat increase.
In other words, exercise increases lean muscle ... and increases BMR.
• Lower your body fat percentage and your get a lower BMR ... fewer calories are burned while resting.
• BMR also depends upon external temperature (on cold days you burn more calories to maintain normal body temperature).
• Internal temperature, too. (You have a cold, you burn more calories). An increase of 4 degrees celsius in body temperature can increase BMR by 50% !!
• Certain glands affect BMR. If you produce too little thyroxin (produced by the thyroid gland) you can get a 30-40% reduction in BMR.
• Diets, fasting, malnutrition ... they all decrease BMR.
>And it depends upon the phases of the moon?
Possibly.
>And you REALLY lost 10 pounds in 10 days?
Yes.
>How'd you do that?
 I fixed a bowl of kidney beans, salt, pepper, garlic powder, oregano, olive oil and vinegar. That's like -- Italian dressing. The bowl sat quietly on the kitchen counter -- with a tablespoon stuck into the creation. Every time I get a little hungry, I'd have a spoonful. Next day the kidney beans are replaced by chick peas or corn and green beans. Maybe I'll toss in pieces of celery or lettuce or shaved carrots or cereal. Maybe I'll replace the Italian dressing with a teablspoon of store-bought salad dressing. ... French or Ranch or Thousand island (where a tablespoon is just 40 calories). Then I munch on pieces of apple, some grapes or cherries, a few nuts ... and drink some fruit juice. By dinner, I'm full -- but manage to finish off the last of the concoction. >Does that diet have a name? Of course! It's the 2N diet ... Nifty Nibbling or maybe the 2G diet ... Graze, not Gorge or maybe ... 10 pounds ... 10 days

>Be serious!
Okay:

... the nibbling plan reduced fasting serum concentrations of total cholesterol about 9%, low-density lipoprotein cholesterol (14%) and apolipoprotein B approximately 15%. Of particular note is the fact that the average insulin level decreased about 28%. In addition, the mean 24-hour urinary cortisol excretion was lowered approximately 17%.
Then there's:
... the mean 24-hour urinary cortisol excretion was lower by 17.3 +/- 5.9 percent (P less than 0.05) at the end of the nibbling diet than at the end of the three-meal diet. The blood glucose, serum insulin, and C-peptide responses to a standardized breakfast and the results of an intravenous glucose-tolerance test conducted at the end of each diet were similar. We conclude that in addition to the amount and type of food eaten, the frequency of meals may be an important determinant of fasting serum lipid levels, possibly in relation to changes in insulin secretion.
 Finally, rats nibble. If a rat is fed just three times a day, the rat becomes obese, the cholesterol rises, and there are blood glucose aberrations of a diabetic nature. >Rats? You're kidding, right? Would I kid you? (Besides, I just write what I read on the Internet. ) Check out Jenkins at my alma mater, Univ. of Toronto. He and his colleagues conducted trials on humans, comparing nibbling to gorging. He was also involved in the discovery of the Glycemic Index. Rats nibble

 Kleiber's law

I recently discovered another interesting thing ... Kleiber's law.
It suggests that the Energy required for mammals to survive is related to their Mass, like so:
 E = k M3/4   where k is some constant
>Mammals? Like you?
Or like you ... or ants or whales or ... whatever. Indeed, there's a suggestion that the law applies to plants, too.

>A cauliflower, perhaps?
Perhaps.
However, some suggest that the exponent should be 2/3.
Nevertheless, here's an guesstimate of the caloric reduction necessary to achieve a given weight loss ... using Kleiber's law.

 Current Weight pounds Desired Weight Loss pounds % calorie reduction
Calculator #5
Calorie reduction to achieve weight loss

 Update
I discovered that the ancient Harris-Benedict Equation (1919) has a reincarnation
... the more recent (1990) Mifflin-St Jeor equation:
 Women: BMR = ( 10 x weight in kg ) + ( 6.25 x height in cm ) - ( 5 x age in years ) - 161 Men: BMR = ( 10 x weight in kg ) + ( 6.25 x height in cm ) - ( 5 x age in years ) + 5

We'll do it in English units, eh?
Notice that the result differs somewhat from good ol' Harris-Benedict.

 Current Weight pounds Height ft   in Age Male or Female ? (0 for Male, 1 for Female) calories per day "at rest" calorie consumption
Remember the magic formula for Equilibrium Weight?
 [3]     Eventual "Equilibrium" Weight = A/B = [ K - f (a + c (height) - d (age) ] / ( f b )
That'd be your "eventual" weight if you continued with a daily consumption of K calories.
If your current weight is W0, then the difference would be:
Eventual Weight - Current Weight = [ K - f (a + c (height) - d (age) ] / ( f b ) - W0 = (K - f BMR0) / fb     where BMR0 is your initial BMR.
 Eventual Weight Change = (K - f BMR0) / fb

To lose weight, we arrange to have K small enough so that (K - f BMR0) < 0, right?
Indeed, the eventual weight loss is proportional to this difference ... regardless of whether we adopt Harris-Benedict or Mifflin-St Jeor.

>And you believe this stuff?
I believe it's interesting, don't you?
>No.
Pay attention.

Suppose we decrease our daily calorie intake by 500 calories, so K - f BMR0 = -500 calories.
Then our eventual weight would decrease (according to the weight gurus) by 500/fb and, picking f = 1.2 (for lethargic lifestyle), that's about = 417/b.
 Aah, but there are different values for b, eh? That means ... >I pick the BMR formula that has the tiniest b-value! Am I right? Undubitably! Everybody knows that it's the fomula that determines your eventual weight ... not your lifestyle. Suppose you reduce your daily consumption by 500 calories (from your current BMR). Then your eventual weight loss will rest upon which formula you adopt >What's the red dots? Them's your choices.
Remember our differential equation?
[2]     dW/dt = A - B W   with solution   W(t) = A/B + (W0 - A/B) e-B t
where:   A = [ K - f (a + c (height) - d (age) ) ] / 3500   and   B = f b / 3500
and f (whose value is between 1 and 2) varies with the amount of your daily activity; lethargic to Olympian.

So (after some fiddling):
 [*]     W(t) = W0 - [(f BMR0 - K) / fb ] (1 - e-fbt/3500)

where:   BMR0 = a + bW0 + c (height) - d (age) is your initial BMR
and a, b, c and d are curious parameters which depend upon whose formula you adopt.

The ubiquitous "3500" calories per pound of body fat is there - in the exponent. Change it a mite and the rate of losing wight can change dramatically.

>You don't believe in the 3500 cals per pound?
I don't believe that, regardless of your genetic configuration, reducing your consumption by 3500 calories in one week (500 cals/day for 7 days) will lose you a pound.
Some people burn calories like crazy, just to stay alive ... others don't. A "universal" 3500 makes little sense.
Look carefully at the equation [*], above, namely: W(t) = W0 - [(f BMR0 - K) / fb ] (1 - e-fbt/3500)
Note that (f BMR0 - K) is your daily calorie reduction. Let's call that ΔC, then we have: W(t) = W0 - (ΔC / fb) (1 - e-fbt/3500).
Now we'll let fb = 3500λ so we have: W(t) = W0 - (ΔC / 3500λ) (1 - e-λt).
We recognize ΔC / 3500 as the number of pounds of body fat which your calorie reduction, ΔC , would give (assuming the ubiquitous "3500" calories per pound).
We then let ΔC / 3500 = β and get (finally!):

 [*]     W(t) = W0 - (β/λ) (1 - e-λt)

See how simple? All the strange parameters associated with Harris-Benedict or Mifflin-St Jeor are buried in the 2 parameters β = ΔC / 3500 and λ = fb / 3500.

>How big are those guys?
Let's see ... for a 1000 daily calorie reduction we'd get β = 1000 / 3500 ... about 0.3.
And λ = fb / 3500 might run in the neighbourhood of 10/3500 or about 0.003. ... and β/λ would be about 100.
Here's a graph of (1 - e-λt)/λ for λ = 0.003.
Of course, for t → ∞, then (1 - e-λt)/λ → 1/λ. (In the diagram, that'd be about 333.)

Here's a "Rule of Thumb" for you:
Assume β is about 100/3500 or about 0.03 for a 100 calorie daily reduction - and take 1/λ = 333.
Then your eventual weight loss is β/λ = 0.03 x 333 ≈ 10 pounds for every 100 calorie daily reduction.
 There's a 10 pound eventual weight reduction for every 100 calorie reduction in daily consumption
>Do you believe that?
A Rule of Thumb: something easy to remember and usually misleading.

However, look at it this way ... assuming the Mifflin-St Jeor formula which has the form:
BMR1 = 10 W1 + (some constant)   where (some constant) don't hardly change unless you get taller or shorter or get older or change sex.
Now suppose that, after dieting for some time you've reached a lower equilibrium weight W2 and your BMR changes to BMR2:
BMR2 = 10 W2 + (some constant)
Subtract and get:
BMR1 - BMR2 = 10 (W1 - W2)
See? Your weight change would be 1/10 of the change in BMR.

>Are you still trying to lose weight? If so, where's the spreadsheet?
Yes, I'm still trying and the spreadsheet is here:

There's an Explain sheet, like so:

It also has a sheet with some food nutrient data, like so:

>How many food items in that list?
A couple of thousand, but here's a spreadsheet with close to 8000 items (where you type in a couple of words to describe the food)::

with an explanation like so:

P.S.
The spreadsheet(s) may change from week to weak ... and therefore may differ from the pretty pictures above.

Also, take a peek at this: the Math of Weight Loss

 Final thoughts

 If you eat a pound of food, you immediately gain a pound. There's no getting away from that fact. That goes for a pound of water or a pound's worth of chocolate milkshake or bacon fat. What happens to that pound over time? That'll depend upon what food you ate, your metabolism, the phases of the moon and the precession of the orbit of Mercury. Every snack, every bite of apple, every cookie ... that'll increase your weight. After eating, if you just breath, walk about and eat nothing, your weight will decrease. (It takes energy just to stay alive, eh?) So eat less, eat some foods which your body refuses to turn into body weight ... and pray that the decreases between meals is greater than the increases at mealtime.
I weighed myself the other day. Then I drank a large glass of tomato juice. Then I weighed myself ag'in.
I had gained one pound! I weighed a large glass of tomato juice. Yes .. it weighed exactly one pound.