# How much glue?

#### oxicottin

##### New member
Hello everyone, This is my first post and im not very good at math and I have been trying to figure out how many bags of glue we go through or we will need.

Here is the glue we use:

UNI-REZ™ 2635 A resin with chemical and grease resistance, low temperature flexibility, excellent high/low temperature recycling, good adhesion to plastics, suitable for cabinet and furniture assembly, multipurpose assembly and adhesion to preheated metals applications.

Ok, now we run the thickness at .0015 at a width of 2.125 and the length is in linier foot lets say 216,000lf. Now our bags come in 50lb bags. How in world do I figure out how much glue we can get out of a bag for the 216,000 linier foot. I have bee reading and im hearing volocity and volume ect. and im confused on what to even calculate and how. "The glue is in beads and melted down"

Thanks!!!

#### MarkFL

Staff member
We can figure this out if we have one other vital piece of information...the weight density of the glue. We need to know how much space a pound of the glue occupies (or be able to compute this if the weight density is given in other units). Is this information given?

#### oxicottin

##### New member
We can figure this out if we have one other vital piece of information...the weight density of the glue. We need to know how much space a pound of the glue occupies (or be able to compute this if the weight density is given in other units). Is this information given?

I will do some digging around and see if I cant find the weight density in the meantime,
How is this figured out anyway? Thanks

OK, I cant find anything on Weight Density so I will have to call the company tomorrow, But here is specks I did find.

Appearance/state: Amber solid
Odor: Mild amine
Ph: Not applicable
Vapor Pressure: <0.001 mm Hg at 20°C (68°F)
Vapor density: Not applicable
Boiling point: Not applicable
Melting point: 140°C
Solubility in water: <0.1% at 25°C (77°F)
Specific gravity: 0.96 at 25°C/25°C (water=1.00)
Evaporation rate: Approx. 0 (n-BuAc=1)
% Volatile by weight: 0

Softening Point, °C, Ring & Ball, ASTM E 28 135-145
Viscosity, LVT Brookfield, Spindle # 31, cps/mPa×s at 190°C, ASTM D 3236 3000-4700
Color, 40% in n-Butanol, Gardner 7 Maximum
TYPICAL VALUES
Acid Value, mg KOH/g 0.9
Amine Value, mg KOH/g 7.5
Color, 40% in n-Butanol, Gardner 5
Tensile Strength, psi (MPa), ASTM D 1708 280 (1.9)
Elongation, %, ASTM D-1708 200
Tensile Modulus, psi (MPa), ASTM D 638 3000 (21)
Peel Strength, pli (N/25mm), plasticized PVC to itself, ASTM D 1876 20 (88)
Mandrel Bend Flexibility, (1” diameter), pass, °C, ASTM D 3111 -30
Open Time, sec., 10 mil film, ASTM D 4497 25
Shear Adhesion Failure Temperature (SAFT), 1 kg. wt. °C, ASTM D 4498 130
Water Absorption, ASTM D-570:
after 24 hours, %
after 7 days, %
1.70
3.11
Pounds per Gallon, (g/cc), 25°C 8.0 (0.96)
Flash Point,°C, COC 271 (520°F)

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#### oxicottin

##### New member
Is this what your looking for?

Pounds per Gallon, (g/cc), 25°C 8.0 (0.96)

Thanks!

#### MarkFL

Staff member
I am assuming you are trying to figure this our for a real world application rather than asking for help on a homework problem, so I will work out a full solution for you.

Given the specific gravity is 0.96, we then know that 1 cubic centimeter is 0.96 grams. Now, there are 231 cubic inches in a gallon and 2.54 centimeters per inch and 453.592 grams per pound. Hence:

$$\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\cdot\frac{1\text{ lb}}{453.592\text{ g}}\cdot\left(\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{12\text{ in}}{1\text{ ft}} \right)^3\approx60\frac{\text{lb}}{\text{ft}^3}$$

Next, we need to compute the volume of glue we need for the given job. Assuming the dimensions you have given are all in ft., this is:

$$\displaystyle V=2.125\,\times\,216000\,\times\,0.0015=688.5\text{ ft}^3$$

This means we need, in pounds of glue:

$$\displaystyle 60\frac{\text{lb}}{\text{ft}^3}\cdot688.5\text{ ft}^3\approx41310\text{ lb}$$.

Dividing this by 50, we find that this equates to about 826.2 50 lb. bags of glue that will be needed for the job.

#### oxicottin

##### New member
Yes its for real word... yes its in inch. was post #4 the weight density you were looking for?

Thanks

#### MarkFL

Staff member
So, the thickness and width are in inches? In that case divide the results I gave above by $12^2=144$ to get $5.7375$ 50 lb. bags.

The specific gravity gave us a means of finding the pounds per cubic foot.

#### oxicottin

##### New member
Given the specific gravity is 0.96, we then know that 1 cubic centimeter is 0.96 grams. Now, there are 231 cubic inches in a gallon and 2.54 centimeters per inch and 453.592 grams per pound. Hence:

$$\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\cdot\frac{1\text{ lb}}{453.592\text{ g}}\cdot\left(\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{12\text{ in}}{1\text{ ft}} \right)^3\approx60\frac{\text{lb}}{\text{ft}^3}$$
.

Mark, Instead the formula above can you write it out in "General Math" as in:

0.096 * this / this = this

I'm sorry I never took algebra or any of those classes (Never had to years ago) so I'm lost on your formula. Thanks!

#### Ackbach

##### Indicium Physicus
Staff member
Realistically, you're not going to get $100\%$ yield from each bag or box of glue. I don't know whether that's going to be an important factor or not. It might be second-order, or even third-order, in which case you might not be interested. As the surface area-to-volume ratio of your bags increases, this effect will increase.

#### oxicottin

##### New member
Realistically, you're not going to get $100\%$ yield from each bag or box of glue. I don't know whether that's going to be an important factor or not. It might be second-order, or even third-order, in which case you might not be interested. As the surface area-to-volume ratio of your bags increases, this effect will increase.
Yes I understand that is not aprox but that still doesnt answer my question in post 9 "how is it calculated in general terms"

Thanks!

#### MarkFL

Staff member
Mark, Instead the formula above can you write it out in "General Math" as in:

0.096 * this / this = this

I'm sorry I never took algebra or any of those classes (Never had to years ago) so I'm lost on your formula. Thanks!
What I did was to take the given specific gravity of $$\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}$$ and then multiplied by 1 in the form of a string of equivalences such that that all units cancel except for the desired units of $$\displaystyle \frac{\text{lb}}{\text{ft}^3}$$.

#### oxicottin

##### New member
What I did was to take the given specific gravity of $$\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}$$ and then multiplied by 1 in the form of a string of equivalences such that that all units cancel except for the desired units of $$\displaystyle \frac{\text{lb}}{\text{ft}^3}$$.
Thank you for the help.... Is this correct then?

0.96 g/cm^3 X 62.4280115 = ___lb/ft^3

Code:
DENSITY
g/cm3. Multiply By  62.43 To Obtain lb/ft3
0.03613 To Obtain lb/in3
Conversion Chart Found Here..

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