Fonz
Elite Mentor,
Author : Fonz
Summary : Mathematical Analysis of blood flow and glucose flow and what haoppens to the ingested glucose.
Glucose Utilization Article (By Fonz) Part I
This is a bit complicated, but please bear with me.
Imagine you consume a meal with carbohydrates, fats, and proteins. So, the original polynomial equation becomes, for the total ingested glucose load becomes:
Glucose Ingested =
(Grams of Carbohydrates) * (1.0) + (Grams of Protein * (0.58) + (Grams of Fat * (0.1)
The equation can be explained in this fashion:
- Carbohydrates convert to glucose with 100% efficiency
- Protein converts to glucose with approx. 58% efficiency because
there are glucogenic and ketogenic amino-acids. (Keep in mind
that glutamine is glucogenic and comprises close to 61% of the
amino-acid pool in the blood stream).
-Fat converts to glucose at a 10% conversion rate due to the
glycerol chain at the end of the triglyceride(fat) molecule.
This equation just represents the total glucose load ingested by a person during a meal containing all macro-nutrients, Fats, proteins, and carbohydrates.
EXCERPT OF MY GLUCOMETRIC ANALYSIS
I’ll now take a small excerpt from my Glucometric Analysis, which you can find at CEM and at EF.
Standardized Meal(Bread) = 500Kcal 6g Fat 14.3g Protein 98g Carbs
According to the equation represented above:
Total Glucose derived from ingested meal =
(98g carbohydrates * 1.0) + ((14.3g protein * (0.58)) + (6g fat * (0.1)) = 98g + 8.29g + 0.6g = 106.89g total glucose ingested.
Supplement #1: Placebo
Person was me(Fonz).
Weight: 82Kg(180.8lbs)
Blood Volume(Explained below) = 6200ml
(T = 0 hrs)
Initial BG Measurement: 48mg/dl Temp:37.3C (99.1F)
(Eat Food as described above)
(T+1hr) Measurement: 90mg/dl Temp: 37.2C (99F)
(T+2hrs) Measurement: 40mg/dl Temp: 36.8C (98.2F)
(T+3hrs) measurement: 74mg/dl Temp: 37.1C (98.8F)
(T+4hrs) Measurement: 72mg/dl Temp: 37.2F (99F)
Area under positive BG Curve (Taking initial BG measurement as the horizontal) =
21 + 17.64 + 11.47 + 24 + 1 = 75.11 mg/dl (squared)
Area under initial BG measurement (negative):0.64 + 0.47 = 1. 11 mg/dl (squared)
Therefore the Total Area = 75.11 + 1.11 = 76.22 mg/dl (squared)
The total Area under the Blood Glucose Curve was 76.22 mg/dl (squared) after the standardized meal. A normal person of 70Kg has 5000ml of blood volume. The blood volume then increases in increments of 0.6L per 6Kg of body weight. So, since I was the test subject, my standard weight was 82Kg(180.8lbs), and had therefore 6.2L or 6200ml of blood. So therefore, I had 76.22mg of glucose per dl(squared) in my blood in any given cross-sectional area of my blood vessels.
Here is where it gets interesting:
Blood Flow Analysis:
Since a blood vessel has a cross-sectional area, 1 decilitre = 0.1 Litres. This is the latter amount of fluid/blood which travels through the cross-sectional of the blood vessel. Finding out the flow rate of the cross-sectional area of the given blood vessel flow can be done this way: Flow Rate = 1 decilitre = 0.1L = 100g(Assume that the density of blood is very close to 1g/ml) Therefore, The Area of the blood vessel = (Pi * Radius(squared)). Taking this into account, the Mass Flow-rate = 100g * Area of the blood vessel. Then, we can surmise that the Mass Flow-rate = 100g * Pi * radius of cross-sectional blood vessel(squared). From reference books one can determine that the average sized blood vessel that carries blood, oxygen and nutrients etc… to parts of the human anatomy in any normal healthy person can be surmised to be approx: 3.5cm in diameter.
Now, one can derrive the radius of the cross-sectional area of the blood vessel where the volume of blood flow saturated with the ingested glucose is flowing. This number is approximately 1.75cm.. Note: The measurement by the blood glucose machine is an area because the volumetric flow rate equation inside a fluidic cross-sectional area is due to the fact that the blood vessel is a circular area(like a hose) not a straight line. If one where to use their imagination think of a line of bread crumbs(glucose in this case) going through a circular hose with a specific diameter. Now, the cross-sectional area of a blood vessel can be calculated as (A) = Pi * Radius(squared) . Since we can take the radius of an average sized blood vessel of a healthy male to be 1.75cm(See above). Then, the cross-sectional area of the blood vessel becomes: Cross-sectional Area of Blood vessel = Pi * ( 1.75cm)(1.75cm) = 9.6211 cm(squared).
Calculations:
So, the total area under the blood glucose curve as seen in the excerpt of my old glucometric analysis above is 76.22mg/dl(squared). But as a decilitre = 0.1L. 0.1L(squared) = 10000ml(squared). 10000ml(squared) being equal to 100 decilitres. From this, the new data becomes that glucose travels at 76.22mg per 10000ml of blood(which is equal to 100dl or 10L) after the ingestion of my aforementioned glucose load.. But, since my total blood volume was 6.2L(See the excerpt of my glucometric analysis) we can now surmise that the blood loaded with glucose goes through blood vessels with a cross-sectional area of approximately 9.6211cm(squared). Therefore, the total glucose travelling through the blood vessels of the body approximately after the initial glucose load is (76.22mg glucose per 10 Litres of blood). In 6.2L of blood(6200000mg), the amount of glucose present in the blood would be 76.22mg * (6.2L/10L)) = 47.2564mg of total glucose flowing around my entire blood plasma vessel structure. But now, we have to do the last part, and that is factor in the cross-sectional area of the blood vessels. From the blood flow analysis section: Mass flow-rate across a circular cross-section of a blood vessel = Mass(glucose(mg)) * Area(cm(squared)).
Therefore, Mass flow-rate = 47.2564mg (glucose) * 9.6211 cm(squared)
But now, we have to change dimensions:
a) 0.0472564g = 47.2564mg
b) 1 cm (squared) = 1.0 * 10exp(-4) metres squared.
Then, 1cm(squared) = 1.0 * 10exp(-6) metres cubed.
Therefore, 9.6211 cm(squared) = 9.6211 * 1.0 * 10exp(-4) metres squared = 0.0009621 metres squared.
Therefore ultimately, 9.6211cm(squared) = 9.6211 * 1.0 * 10exp(-6) = 0.000009621 metres cubed.
Now using the dimensions we have derived previously:
(From my previous Glucometric Analysis)
Standardized Meal(Bread) = 500Kcal 6g Fat 14.3g Protein 98g Carbs
According to the equation I described above:
Total Glucose derived from ingested meal = 98g + 8.29g + 0.6g = 106.89g total glucose
Total glucose inside my blood supply after initial glucose load was: 47.2564mg of glucose. Total cross-sectional Area of an average blood vessel in cubic metres = 0.000009621. Now, one metre cubed = 1000Litres, therefore 0.000009621 metres cubed = 0.009621L. Which this can then be changed to decilitre format, or 0.09621dl. So, therefore total linear area of blood is 0.09621dl And the glucose mass was 47.2564mg
Therefore, we are ready for the final phase.
Total Glucose content = 47.2564mg
Total Area Blood = 0.09621 decilitres.
But, we have to reduce the final equation to the standard mg/dl to compare glucose variances and losses. Therefore, the final standard blood glucose measurement through a blood vessel becomes (47.2564mg/0.09621dl). Which if reduced proportionally in order for dl = 1(constant), you get: BG(measurement) for the entire body = (47.2564*10.3939)mg/dl = 491.178mg/dl
Now, the glucose load(initial) was 106.89g = 106890mg
Amount of blood plasma in my body = 6.2L = 62dl
Therefore, the BG measurement(whole body should read)(106890mg/62dl) = 1724.03mg/dl
Discussion:
There seems to be a discrepancy between the blood glucose level in the blood supply T=4hrs after ingestion of the standardized meal(491.178mg/dl), and what was measured by the initial glucose load after ingestion of the standardized meal( 1724.03mg/dl). This seems to suggest that the orally ingested macro-nutrient meal described beforehand of a specific glucose load, was then diverted to the different organs of the body, - such as the muscle cells, fat cells, liver cells, oxidized for fuel, brain cells, etc… This I believe can be explained by the differential in the initial glucose load per unit blood after the ingestion of the original meal(in decilitres): 1724.03mg/dl, and then the subsequent drop in measured blood glucose levels afterwards(in decilitres): 491.178mg/dl.
These two numbers: Initial: 1724.03mg/dl and Final: 491.178mg/dl, indicate that (100% - (491.178/1724.03) * 100%) = 71.51% of the original glucose load was used by the body for various uses. The rest was lost to unknown variances. Converting it to grams, we get:
Initial load: 106.89g Glucose Final(measured) in blood vessels: 30.453g Glucose
The discrepancy being (106.89g – 30.453g) = 76.437g. Therefore, only 76.437g of glucose out of 106.89g of ingested glucose was used by the body under normal conditions.
This represents a rather curious discrepancy.
Author's Comments : It is more of a Part I article. part II will be far more complex. To write a mathematical formula that will predict where exactly the glucose that was ingested actually goes..albeit the fat, muscle, liver, cells, oxidized, brain, etc...
Fonz
Summary : Mathematical Analysis of blood flow and glucose flow and what haoppens to the ingested glucose.
Glucose Utilization Article (By Fonz) Part I
This is a bit complicated, but please bear with me.
Imagine you consume a meal with carbohydrates, fats, and proteins. So, the original polynomial equation becomes, for the total ingested glucose load becomes:
Glucose Ingested =
(Grams of Carbohydrates) * (1.0) + (Grams of Protein * (0.58) + (Grams of Fat * (0.1)
The equation can be explained in this fashion:
- Carbohydrates convert to glucose with 100% efficiency
- Protein converts to glucose with approx. 58% efficiency because
there are glucogenic and ketogenic amino-acids. (Keep in mind
that glutamine is glucogenic and comprises close to 61% of the
amino-acid pool in the blood stream).
-Fat converts to glucose at a 10% conversion rate due to the
glycerol chain at the end of the triglyceride(fat) molecule.
This equation just represents the total glucose load ingested by a person during a meal containing all macro-nutrients, Fats, proteins, and carbohydrates.
EXCERPT OF MY GLUCOMETRIC ANALYSIS
I’ll now take a small excerpt from my Glucometric Analysis, which you can find at CEM and at EF.
Standardized Meal(Bread) = 500Kcal 6g Fat 14.3g Protein 98g Carbs
According to the equation represented above:
Total Glucose derived from ingested meal =
(98g carbohydrates * 1.0) + ((14.3g protein * (0.58)) + (6g fat * (0.1)) = 98g + 8.29g + 0.6g = 106.89g total glucose ingested.
Supplement #1: Placebo
Person was me(Fonz).
Weight: 82Kg(180.8lbs)
Blood Volume(Explained below) = 6200ml
(T = 0 hrs)
Initial BG Measurement: 48mg/dl Temp:37.3C (99.1F)
(Eat Food as described above)
(T+1hr) Measurement: 90mg/dl Temp: 37.2C (99F)
(T+2hrs) Measurement: 40mg/dl Temp: 36.8C (98.2F)
(T+3hrs) measurement: 74mg/dl Temp: 37.1C (98.8F)
(T+4hrs) Measurement: 72mg/dl Temp: 37.2F (99F)
Area under positive BG Curve (Taking initial BG measurement as the horizontal) =
21 + 17.64 + 11.47 + 24 + 1 = 75.11 mg/dl (squared)
Area under initial BG measurement (negative):0.64 + 0.47 = 1. 11 mg/dl (squared)
Therefore the Total Area = 75.11 + 1.11 = 76.22 mg/dl (squared)
The total Area under the Blood Glucose Curve was 76.22 mg/dl (squared) after the standardized meal. A normal person of 70Kg has 5000ml of blood volume. The blood volume then increases in increments of 0.6L per 6Kg of body weight. So, since I was the test subject, my standard weight was 82Kg(180.8lbs), and had therefore 6.2L or 6200ml of blood. So therefore, I had 76.22mg of glucose per dl(squared) in my blood in any given cross-sectional area of my blood vessels.
Here is where it gets interesting:
Blood Flow Analysis:
Since a blood vessel has a cross-sectional area, 1 decilitre = 0.1 Litres. This is the latter amount of fluid/blood which travels through the cross-sectional of the blood vessel. Finding out the flow rate of the cross-sectional area of the given blood vessel flow can be done this way: Flow Rate = 1 decilitre = 0.1L = 100g(Assume that the density of blood is very close to 1g/ml) Therefore, The Area of the blood vessel = (Pi * Radius(squared)). Taking this into account, the Mass Flow-rate = 100g * Area of the blood vessel. Then, we can surmise that the Mass Flow-rate = 100g * Pi * radius of cross-sectional blood vessel(squared). From reference books one can determine that the average sized blood vessel that carries blood, oxygen and nutrients etc… to parts of the human anatomy in any normal healthy person can be surmised to be approx: 3.5cm in diameter.
Now, one can derrive the radius of the cross-sectional area of the blood vessel where the volume of blood flow saturated with the ingested glucose is flowing. This number is approximately 1.75cm.. Note: The measurement by the blood glucose machine is an area because the volumetric flow rate equation inside a fluidic cross-sectional area is due to the fact that the blood vessel is a circular area(like a hose) not a straight line. If one where to use their imagination think of a line of bread crumbs(glucose in this case) going through a circular hose with a specific diameter. Now, the cross-sectional area of a blood vessel can be calculated as (A) = Pi * Radius(squared) . Since we can take the radius of an average sized blood vessel of a healthy male to be 1.75cm(See above). Then, the cross-sectional area of the blood vessel becomes: Cross-sectional Area of Blood vessel = Pi * ( 1.75cm)(1.75cm) = 9.6211 cm(squared).
Calculations:
So, the total area under the blood glucose curve as seen in the excerpt of my old glucometric analysis above is 76.22mg/dl(squared). But as a decilitre = 0.1L. 0.1L(squared) = 10000ml(squared). 10000ml(squared) being equal to 100 decilitres. From this, the new data becomes that glucose travels at 76.22mg per 10000ml of blood(which is equal to 100dl or 10L) after the ingestion of my aforementioned glucose load.. But, since my total blood volume was 6.2L(See the excerpt of my glucometric analysis) we can now surmise that the blood loaded with glucose goes through blood vessels with a cross-sectional area of approximately 9.6211cm(squared). Therefore, the total glucose travelling through the blood vessels of the body approximately after the initial glucose load is (76.22mg glucose per 10 Litres of blood). In 6.2L of blood(6200000mg), the amount of glucose present in the blood would be 76.22mg * (6.2L/10L)) = 47.2564mg of total glucose flowing around my entire blood plasma vessel structure. But now, we have to do the last part, and that is factor in the cross-sectional area of the blood vessels. From the blood flow analysis section: Mass flow-rate across a circular cross-section of a blood vessel = Mass(glucose(mg)) * Area(cm(squared)).
Therefore, Mass flow-rate = 47.2564mg (glucose) * 9.6211 cm(squared)
But now, we have to change dimensions:
a) 0.0472564g = 47.2564mg
b) 1 cm (squared) = 1.0 * 10exp(-4) metres squared.
Then, 1cm(squared) = 1.0 * 10exp(-6) metres cubed.
Therefore, 9.6211 cm(squared) = 9.6211 * 1.0 * 10exp(-4) metres squared = 0.0009621 metres squared.
Therefore ultimately, 9.6211cm(squared) = 9.6211 * 1.0 * 10exp(-6) = 0.000009621 metres cubed.
Now using the dimensions we have derived previously:
(From my previous Glucometric Analysis)
Standardized Meal(Bread) = 500Kcal 6g Fat 14.3g Protein 98g Carbs
According to the equation I described above:
Total Glucose derived from ingested meal = 98g + 8.29g + 0.6g = 106.89g total glucose
Total glucose inside my blood supply after initial glucose load was: 47.2564mg of glucose. Total cross-sectional Area of an average blood vessel in cubic metres = 0.000009621. Now, one metre cubed = 1000Litres, therefore 0.000009621 metres cubed = 0.009621L. Which this can then be changed to decilitre format, or 0.09621dl. So, therefore total linear area of blood is 0.09621dl And the glucose mass was 47.2564mg
Therefore, we are ready for the final phase.
Total Glucose content = 47.2564mg
Total Area Blood = 0.09621 decilitres.
But, we have to reduce the final equation to the standard mg/dl to compare glucose variances and losses. Therefore, the final standard blood glucose measurement through a blood vessel becomes (47.2564mg/0.09621dl). Which if reduced proportionally in order for dl = 1(constant), you get: BG(measurement) for the entire body = (47.2564*10.3939)mg/dl = 491.178mg/dl
Now, the glucose load(initial) was 106.89g = 106890mg
Amount of blood plasma in my body = 6.2L = 62dl
Therefore, the BG measurement(whole body should read)(106890mg/62dl) = 1724.03mg/dl
Discussion:
There seems to be a discrepancy between the blood glucose level in the blood supply T=4hrs after ingestion of the standardized meal(491.178mg/dl), and what was measured by the initial glucose load after ingestion of the standardized meal( 1724.03mg/dl). This seems to suggest that the orally ingested macro-nutrient meal described beforehand of a specific glucose load, was then diverted to the different organs of the body, - such as the muscle cells, fat cells, liver cells, oxidized for fuel, brain cells, etc… This I believe can be explained by the differential in the initial glucose load per unit blood after the ingestion of the original meal(in decilitres): 1724.03mg/dl, and then the subsequent drop in measured blood glucose levels afterwards(in decilitres): 491.178mg/dl.
These two numbers: Initial: 1724.03mg/dl and Final: 491.178mg/dl, indicate that (100% - (491.178/1724.03) * 100%) = 71.51% of the original glucose load was used by the body for various uses. The rest was lost to unknown variances. Converting it to grams, we get:
Initial load: 106.89g Glucose Final(measured) in blood vessels: 30.453g Glucose
The discrepancy being (106.89g – 30.453g) = 76.437g. Therefore, only 76.437g of glucose out of 106.89g of ingested glucose was used by the body under normal conditions.
This represents a rather curious discrepancy.
Author's Comments : It is more of a Part I article. part II will be far more complex. To write a mathematical formula that will predict where exactly the glucose that was ingested actually goes..albeit the fat, muscle, liver, cells, oxidized, brain, etc...
Fonz
