Formula for Masonry Construction
When estimating the amount of masonry units you need for a project you first need to know the square foot coverage and then multiply that by the units per square foot. To Find the amount of certain masonry units as they are called, such as brick or block etc. it is good to know their common averages per Sq. Ft. To find:
# of standard size brick per Sq. Ft. of wall:
# of Queen or Jumbo size brick per Sq. Ft. of wall:
# of 8"x 16" face size CUM (Concrete Block) per Sq. Ft. of wall:
# of Queen or Jumbo size brick per Sq. Ft. of wall:
# of 8"x 16" face size CUM (Concrete Block) per Sq. Ft. of wall:
Multiply Sq. Ft. of wall by (7.25).
Multiply Sq. Ft. of wall by (6.75).
Multiply Sq. Ft. of wall by (1.125)
Multiply Sq. Ft. of wall by (6.75).
Multiply Sq. Ft. of wall by (1.125)
For instance, the face size in square inches of an 8"x 8"x 16" CMU also called an 8" block is 8 x 16, or 128" in. So a block is 128/144 of a Sq. Ft. or 0.8889 Sq.Ft. And there are 144/128 block per square foot of wall, which ='s 1.125.
How to lay out square by the Phythagoras Theorem
When laying out a wall (A) that needs to be square with another (B), it is important to measure the diagonal (C) between the end points of (A) and (B) and then to use a "range" line to align the walls, point to point. This is done by knowing the lengths between points. Useing the Pythagoras theorem is the best way to do this. If wall (A) is 25' and wall (B) is going to be 40', then the "Diagonal" (C) will be equal to the square root of (A) squared,= (A x A) + (B) squared,= (B x B), or 25^2 + 40^2 = (C)^2. Using a scientific calculator it is easy to find (C), but if you don;t have one on site there is another way to find (C). We will cover that later.
OK let's continue; (A)^2= 25^2 = 625, and (B)^2= 40^2 = 1600, the total of (A)^2 and (B)^2 = 625 + 1600 = 2225 so then, the "Diagonal" legnth is the square root of 2225. The square root of 2225 = 47.17, and this is equivalant to 47' and 0.17 or 17/100 of a foot. 17/100 of a foot is about = 2". So the diagonal measurement between (A) and (B) shoiuld be
47' ft. and 2" in.
OK let's continue; (A)^2= 25^2 = 625, and (B)^2= 40^2 = 1600, the total of (A)^2 and (B)^2 = 625 + 1600 = 2225 so then, the "Diagonal" legnth is the square root of 2225. The square root of 2225 = 47.17, and this is equivalant to 47' and 0.17 or 17/100 of a foot. 17/100 of a foot is about = 2". So the diagonal measurement between (A) and (B) shoiuld be
47' ft. and 2" in.
Long division method to find the square root of a number.
To know the square root of 2225 and you don't have a calculater follow these steps:
First: mark off two places at a time from the decimal going to the right and left of the decimal:
= 22 , 25 . 00,00
Second: figure the highest whole number that can be squared as close as possible to 22, the first pair of numbers in 22,25.00,00 This is 4, because 4^2 = 16
5 is too high because 5^2 = 25.
so the first integer in the root of 2225 is 4: with a remainder of 6.
Third: take next piar of numbers (25) and put it with the remainder 6 to make 625
multiply the first integer of 4 by 20 : 4 x 20 = 80
and figure how many times 80 + (X; the number that 80 factors into 625) and multiply that by (X)
In this case, x = 7 because (80 + 7) times 7 = 609. We cannot use 8 for (X) because that would equal
(80 + 8) times 8 = 704, too high for 625.
So the first two numbers in the diagonal are 47. We see that this is correct with the drawing and calculater.
The remainder is 625 - 609 = 16. bring down the pair of zero after the decimal of 2225.00,00 and put them at the end of the remainder of 16. producing 1600
Multiply the Quotient, (as we will call it here) of 47 from above times 20 to get = 940. And again, this time useing the 940, figure (940 + (X)) times (X) to get as close to 1600 as possible.
One time is all that will work so (x) = (1): , (940 + 1) x 1 = 941.
then as above: 1600 - 941 = 659
Bring down the other pair of zero to the end of this remainder: 65900.
For our quotient we have (47.1). omitting the decimal, Multiply 471 by 20 which = 9420.
(9420 + (x)) times (x) to get as close to 65900. It is (6.9) or (7). (x) = 6.9
6.9 x 9426.9 = 65045.61
65900 - 65046 = 854, So we see that (x)= 6. 9 and we will round that to (7).
This proccess can go on and on but I stop when I have rounded off to the second pair of numbers to the left of the decimal in most cases.
First: mark off two places at a time from the decimal going to the right and left of the decimal:
= 22 , 25 . 00,00
Second: figure the highest whole number that can be squared as close as possible to 22, the first pair of numbers in 22,25.00,00 This is 4, because 4^2 = 16
5 is too high because 5^2 = 25.
so the first integer in the root of 2225 is 4: with a remainder of 6.
Third: take next piar of numbers (25) and put it with the remainder 6 to make 625
multiply the first integer of 4 by 20 : 4 x 20 = 80
and figure how many times 80 + (X; the number that 80 factors into 625) and multiply that by (X)
In this case, x = 7 because (80 + 7) times 7 = 609. We cannot use 8 for (X) because that would equal
(80 + 8) times 8 = 704, too high for 625.
So the first two numbers in the diagonal are 47. We see that this is correct with the drawing and calculater.
The remainder is 625 - 609 = 16. bring down the pair of zero after the decimal of 2225.00,00 and put them at the end of the remainder of 16. producing 1600
Multiply the Quotient, (as we will call it here) of 47 from above times 20 to get = 940. And again, this time useing the 940, figure (940 + (X)) times (X) to get as close to 1600 as possible.
One time is all that will work so (x) = (1): , (940 + 1) x 1 = 941.
then as above: 1600 - 941 = 659
Bring down the other pair of zero to the end of this remainder: 65900.
For our quotient we have (47.1). omitting the decimal, Multiply 471 by 20 which = 9420.
(9420 + (x)) times (x) to get as close to 65900. It is (6.9) or (7). (x) = 6.9
6.9 x 9426.9 = 65045.61
65900 - 65046 = 854, So we see that (x)= 6. 9 and we will round that to (7).
This proccess can go on and on but I stop when I have rounded off to the second pair of numbers to the left of the decimal in most cases.
I would agree that this seems to be a complicated proccess, but after you work it a few times it will come naturally. Kind of like long division. I have posted it on my site because it is very important to be able to know the digonal if you are laying out a larger project that needs to be square. It is not easy to find this formula, and it is complicated at first. Sure, you can in most cases use the 3, 4, 5, rule and then use a range line with a tape measure to find where the corner is, but this is not always the case unless you are on a near perfect construction site and have six hands available. Of course if you have a scaled drawing then a rule can be used to measure the diagonal, but take into consideration "human error" using the ruler. Then with a calculater you might find that it does not have working batteries, or you left it at home, or just plain out lost it, will you stop what your doing and go get another? No, but if you hold this method of finding the square root of a number in your brain it will always be with you. Though remember It is very important to check and re-check your calculations at least three times to be sure of the diagonal (Hypotenuse) measurment. You will save time and money by allowing the job to continue with little hinderance.
