If this "formula" is factual then being a righty or lefty should make no difference, learn to use the tool the way it was designed to be used.

Jake

This particular adaptation of a standard math calculation is my own, developed over 25 yars ago. You won't see it in a book. As the originator, I would have thought I had the liberty to suggest a column reversal for lefties.

"If this "formula" is factual ...."

If you stay in college long enough this subject will be covered. It won't look the same. Them academic folks like to make things look complcated. Makes the students think they're gettin' something for their money.

Some of the more practical schools tone down the BS a bit and teach something that is almost usable. The Paul Smith College in upstate New York has an online page that I have linked to below and it shows an academic approach to the problem.

Near the bottom there are some example problems. Just to keep my skills up, I think I'll work the 5 sided polygon of example 3. Since I'm an uneducated redneck with no book learnin', I'll have to do it my way.

Good field math is that which can be done in the midst of loud heavy equipment when the temp is a 100 and the sweat is running down in your eyes.

There seems to be zero interest in this thread. I'm almost sorry I started it.

Here is a link that might be useful: Paul Smth's College, area by coordinates

pls8xx you are adding an interesting perspective(et boucoup plus de exprience) to this forum, you are doing a good job. pls don't give up.

I'll try the problem tomorrow- I was just breezing by the forum on my way to bed. I'm too fuzzy right now for this type of exercise.

I'm with ink- I'll take any good short cuts/instruction I can get.

It's all about bringing in the project on time, on budget. Doing accurate field computations is a part of keeping control.

In its self, area calculations may not be a big factor in cost control, but when added to other proceedures, it is the heart of determining grade cut and fill quantities, which can have a much greater impact.

I did my homework, the Example 3 refered to above. My worksheet ....

{{gwi:42808}}

The college boys got this one right too. I didn't check their answers to the other problems.

I didn't invent the math. Others use a proceedure similar to mine, like the guys at the Alabama DOT. I link one of their pages below(pdf file, pg 6 of 6). But like me, they sometimes do the computations on a tailgate, not in a classroom. I still like my way better.

I listed the coordinates in the opposite order (maybe because I'm a lefty?). All my values were negatives, but the digits match.

And it is hard to mask the numbers with your dominant hand- I can calculate with the right, but I ended up writing the values from right to left on the left side of the chart. And lifting up from the chart makes it hard to get back to where you were. I'm sure I'll get as good as a righty with time!

Lefties should reverse all the column on the worksheet as below ...

{{gwi:42810}}

Two fingers of the right hand keep the place on the form while the left hand is free to run the calculator.

Many thanks pls8xx for posting this, just what I've been looking for. Have you tried using a GPS receiver to get the coordinates? Set your GPS to read UTM instead of degrees, minutes, and seconds and the last few digits of the readout is in meters east and north of a reference line. Do the calculations and convert back to square feet, yards, furlongs or whatever. Once the WAAS feature is operational this fall the accuracy should be about +/- 6 feet or so. I'm going to try it at the farm when the weather cools off a little bit.

Thanks again,
Ichabod

I understood your instructions for lefties- I just wanted to see how 'just use the tool' would work. I can pour punch out of a right handed ladle, but it is easier for me if it has two lips.

I could use the tool, but it will be easier if the columns are reversed, as you suggested.

Interesting. I really like this. It seams to be useful for straight lines. But keep in mind that simple geometry can be easy too.

I see Area Square(#1,#2,#4,#5) minus Area Triangle(#2,#3,#4).

{{gwi:42812}}

Okay, i give. I've been looking at this thread for awhile and I'm very intimidated by the math and angles! But I'm going to give it a try. I've had many times at a site when I couldn't figure out how to take a measurement correctly. Anyway I'll let you now if I'm successful and have one of those Eureka moments. Trust me, if I can do it, anyone can!

Missing Image (sorry).
{{gwi:42814}}

Another simple geometry solution would be:

Shaded_Area = 3/4 x Area_of_Square

RE: On Site Calculations - Area

• Posted by: ichabod z8GA (My Page) on Fri, Sep 8, 06 at 11:49

If you have a simple area you have a simple solution. I'm trying to find the area of a plot of pine trees that has a shape about like the State of Texas and I have no drawing of it. How would drhooke solve that? I'm going to use my gps to find UTM coordinates and use the method pls8xx proposed.

Ichabod

RE: On Site Calculations - Area

• Posted by: pls8xx z8a AR (My Page) on Fri, Sep 8, 06 at 12:13

As drhooke points out there are a number of simple ways to find the area of the first example I used. Even the irregular polygon I linked to at Paul Smth's College can be subdivided into triangles and then worked by aplying the function 1/2 base * height. The problem is you have to calculate the height of each of the triangles first. My experience is that irregular polygons are common in real projects.

Let's consider what a real life project might be like. Suppose you are well into construction of a re-do of an upscale backyard. The area to have sod has been marked on the ground.

Now it could be that the base map used for design was very precise. Cad was used and the area for sod was accurately provided by the software. There have been no changes of the design in construction and the area for sod has been staked on the ground per the plan. If this is the way all your projects go, then you can skip the rest of this post.

The contract and bid documents list 1200 sq yd of sod. What if the quantity is really a little more or less.

Let's take the 'more' look first. The contractor orders 1200 yd and begins laying at the back of the property so that equipment is not driven over the laid sod. He runs out of sod a 100 yd short of doing the area around the patio.

What are your options. You might try to bluff the contractor by telling him it was a Cad design and the grass farm must have shorted him on the sod. Get caught doing this and you will have big trouble way past this project.

Or you could have the extra sod laid and suffer the cost overrun. The contractor doesn't want to pay the delivery fee for another 100 yd and happens to have some left over sod from three weeks ago, still alive, but barely. More trouble. Maybe you can have him pull up sod at the back of the property and move it to fill in around the patio. Yep, he'll do it as soon as you give him a change order so he gets paid for the work.

Now if you had just made a few measurements and calc'ed the area before the sod was ordered you would have known what the situation was. All of the options would have been open to you. You could go with a cost overrun, srink the sod area in a place you could live with, or take some other solution.

What if the sod area is really less. The contractor orders 1200 yd and has 100 yd left over. The documents said 1200 yd, he ordered 1200. Do you want to try and tell the guy he will only get paid for 1100 yd? But your client only got 1100.

Look how easy it would have been if you had told the contractor before he ordered the sod that there was really only 1100 yd to be done and that would be all that he would be paid for.

Keep things straight and fair for both the client and the contractor. You'll finnish with a happy client and a contractor quick to make a good bid on your next project.

RE: On Site Calculations - Area

• Posted by: cady 6b/Sunset34 MA (My Page) on Fri, Sep 8, 06 at 12:56

For some reason, the images aren't showing up on my view. The tables do, but not the diagrams. Phooey.

But, I remember this calculus both from my landscape architecture textbook, and from an instruction book on building water gardens (for calculating the amount of concrete for an irregular pool, and for the volume of water with which to fill it).

It has been handy for caluculating amounts of peastone and mulch, but I've not had to use it more than a couple times. Usually, the spaces I'm working are conventional rectangles, squares, circles. Dumb luck, most probably!

RE: On Site Calculations - Area

• Posted by: laag z6CapeCod (My Page) on Fri, Sep 8, 06 at 17:33

Areas of organic shapes are easily measured by baseline dimensioning and then averaging. That is running out a tape and taking perpendicular measurements at equal intervals from that baseline. You add up all of the lateral dimensions and divide by the number of those dimensions. Then multiply by the length of the baseline. It is accurate and simple.

When rougher esitmates need to be made quickly on not too wavy an area, you can measure the length of the area and then eyeball where the average width is perpendicular to that length and measure that width. Then multiply by the baseline and you know the approximate area.

RE: On Site Calculations - Area

• Posted by: pls8xx z8a AR (My Page) on Sat, Sep 9, 06 at 20:36

Ichabod,
The GPS should give good results for timber areas over 200 acres. For smaller tracts the percentage of error may be more than you would want.

cady, the diagrams for the tables are not posted here. I'm working an example problem from the Paul Smith College web site. Check the link at the bottom of this post to go there.

laag gives us another way to estimate area. As they say .."More than one way to skin a cat." We might want to look at how this works.

Below is a graphic of an "organic" shape on a grid with a base line shown. Perhaps laag or anyone else could suggest an interval distance to use on this model. Hopefully the grid is clear enough give the base line to shape edge distances. Anyone care to work it with me? Refer to laag's post above for instructions.

{{gwi:42816}}

Here is a link that might be useful: for cady, diagram for example problem

RE: On Site Calculations - Area

• Posted by: cady 6b/Sunset34 MA (My Page) on Sat, Sep 9, 06 at 20:54

Thanks for the link to the example, Pls8xx. That one came through intact.

RE: On Site Calculations - Area

• Posted by: ichabod z8GA (My Page) on Mon, Sep 11, 06 at 11:36

(Quote)The GPS should give good results for timber areas over 200 acres. For smaller tracts the percentage of error may be more than you would want. (Unquote)

Not necessarily. I just "surveyed" my property using my Garmin Map76. To save time on the calculations I used the metric calculator at the link below and got 2.49 Hectares, which is 6.15 acres. My official plat shows it to be 6.46 acres, or less than 5% error. Might have gotten it closer but I have lost two of my corner pins.

I was simply suggesting a quick easy way to get the coordinates in the field. Been wondering, how do you get them?

Ichabod

Here is a link that might be useful: Woodlot calculator

RE: On Site Calculations - Area

• Posted by: pls8xx z8a AR (My Page) on Thu, Sep 14, 06 at 9:26

Ichabod,

My comments on the precision of GPS for small tracts was based on what is generally acceptable in today's world. If you go to a meat market to buy steak you expect the scale used to measure it to be better than plus or minus 5%. Same thing at the gas pump. Most commercial ventures of any type operate on a final bottom line margin of not much more than 10%. Logic would hold that quantity management of any part of the operation is desired to be less than 1 to 2%.

Put yourself in a job, or even a marrige, where the profit margin is 1% and everyone gets stressed out. It ain't fun. I hate accounting, being a bean counter is one small step up from being a ditch digger. They are a neccessary evil to maintain a profit, but they tend to be penny wise and pound foolish. You can't let 'em run the show.

I would expect GPS to get 1% precision on tracts greater than 200 acres. Use of two GPS units and a radio where both units are read at the same time, and coordinates are taken based on the difference in position, should give 1% precision on tracts greater than 40 acres. The expensive differential GPS units that surveyors use will get high precision on tracts as small as 2000 sq ft.

You ask how I would obtain coordinates. It would depend on the cimcumstance. For areas under about 30,000 sq ft I would make taped measurements from two perpendicular base lines. For blocky shaped tracts up to 10 acres I would probably do angle and distance measurements from a central location. Tracts larger than 10 acres and smaller than 100 acres need a survey crew. Above that and GPS takes over.

RE: On Site Calculations - Area

• Posted by: pls8xx z8a AR (My Page) on Fri, Sep 15, 06 at 9:38

When making complex calculations one should always seek to employ some element of redundancy as a check for gross errors and blunders. Look back at the last graphic I posted. If the grid units are in feet it is easy to see that the approximate dimensions are about 180 by 95, or an area of 17,100 sq ft. Any calculation of area that is way off this amount will be wrong.

When I invited others to estimate the area of the shape, I failed to tell you that it is composed of standard geometrical shapes, segments of pure circular curves connected with straight lines. As such, the shape can be subdivided into its components and the true and exact area can be determined. I did this and the true area of the shape is 16,303.5

Then I made estimates of the area by the baseline average method suggested by laag. For the first run I used an interval of 20 feet. The 20 data points used, left and right, were:
(0,0) (0,0) (20,56) (20,-30) (40,59) (40,-49) (60,56) (60,-48) (80,52) (80,-43), (100,53) (100,-43) (120,55) (120,-50) (140,50) (140,-55) (160,41) (160,-48) (180,24) (180,0).

The area derived was 16,240 or 99.6% of the true value.

For a second run I chose an interval of 30 feet.
(0,0) (0,0) (30,56) (30,-40) (60,56) (60,-48) (90,51) (90,-40) (120,55) (120,-50) (150,46) (150,-54) (180,24) (180,0)

The result was 15,606 or 95.7% or the true value.

These two calculations demonstrate what you can expect from the baseline average method of estimating area. The precision is dependant on the shape, the interval selected and how the intervals fit the shape. It can give good results but errors of plus or minus 5% are not uncommon.

Are you wondering what I would have done in the field? In the graphic below I show how I would have walked around the shape and set points such that the straight line between them was a good representation of the boundary.

{{gwi:42818}}

Calculation is done as I outlined at the start of this thread.
The data points are:
(0,0) (15,50) (48,62) (85,50) (115,55), (150,48) (180,25) (180,0) (167.5,-42.5) (150,-55) (135,-55) (90,-40) (45,-52) (19,-30)

The result is 16,270 or 99.8% of the true value. This method always yields high precision values, the measurements required are not more than needed for other methods, and the math is just not that complicated.

RE: On Site Calculations - Area

• Posted by: Debbie 5 (debiluis@hotmail.com) on Mon, May 28, 07 at 16:35

I am trying to figure out the square footage of our property. The measurements are: back= 180.58, left side= 134.71, right side= 163.91, front= 51.21. I know this is an irregular polygon, but I do not know how to calculate this. Please help.

RE: On Site Calculations - Area

• Posted by: pls8xx z8a AR (My Page) on Mon, May 28, 07 at 16:52

The calculation cannot be done from the info given.

I am going to guess that the measurements were taken from some type of survey plat. Look and see if directions of the lines are shown (such as N23 47' W), or maybe angles at the line intersections are shown.

The calculation can be done with the addition on one angle or the direction of two connecting lines.

RE: On Site Calculations - Area

• Posted by: saxp (My Page) on Fri, Jan 25, 08 at 7:40

This is a version of Surveyors formula:
Area = [(X1*Y2) - (X2*Y1)] + ... + [(Xn*Y1) - (X1*Yn)]

There are some software that does just that. I use ArePoly (www.openspacesoft.com), it's simple and quick.

Here is a link that might be useful: Arepoly - Open space software

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{{gwi:42814}}

Another simple geometry solution would be:

Shaded_Area = 3/4 x Area_of_Square

If you have a simple area you have a simple solution. I'm trying to find the area of a plot of pine trees that has a shape about like the State of Texas and I have no drawing of it. How would drhooke solve that? I'm going to use my gps to find UTM coordinates and use the method pls8xx proposed.

Ichabod

As drhooke points out there are a number of simple ways to find the area of the first example I used. Even the irregular polygon I linked to at Paul Smth's College can be subdivided into triangles and then worked by aplying the function 1/2 base * height. The problem is you have to calculate the height of each of the triangles first. My experience is that irregular polygons are common in real projects.

Let's consider what a real life project might be like. Suppose you are well into construction of a re-do of an upscale backyard. The area to have sod has been marked on the ground.

Now it could be that the base map used for design was very precise. Cad was used and the area for sod was accurately provided by the software. There have been no changes of the design in construction and the area for sod has been staked on the ground per the plan. If this is the way all your projects go, then you can skip the rest of this post.

The contract and bid documents list 1200 sq yd of sod. What if the quantity is really a little more or less.

Let's take the 'more' look first. The contractor orders 1200 yd and begins laying at the back of the property so that equipment is not driven over the laid sod. He runs out of sod a 100 yd short of doing the area around the patio.

What are your options. You might try to bluff the contractor by telling him it was a Cad design and the grass farm must have shorted him on the sod. Get caught doing this and you will have big trouble way past this project.

Or you could have the extra sod laid and suffer the cost overrun. The contractor doesn't want to pay the delivery fee for another 100 yd and happens to have some left over sod from three weeks ago, still alive, but barely. More trouble. Maybe you can have him pull up sod at the back of the property and move it to fill in around the patio. Yep, he'll do it as soon as you give him a change order so he gets paid for the work.

Now if you had just made a few measurements and calc'ed the area before the sod was ordered you would have known what the situation was. All of the options would have been open to you. You could go with a cost overrun, srink the sod area in a place you could live with, or take some other solution.

What if the sod area is really less. The contractor orders 1200 yd and has 100 yd left over. The documents said 1200 yd, he ordered 1200. Do you want to try and tell the guy he will only get paid for 1100 yd? But your client only got 1100.

Look how easy it would have been if you had told the contractor before he ordered the sod that there was really only 1100 yd to be done and that would be all that he would be paid for.

Keep things straight and fair for both the client and the contractor. You'll finnish with a happy client and a contractor quick to make a good bid on your next project.

For some reason, the images aren't showing up on my view. The tables do, but not the diagrams. Phooey.

But, I remember this calculus both from my landscape architecture textbook, and from an instruction book on building water gardens (for calculating the amount of concrete for an irregular pool, and for the volume of water with which to fill it).

It has been handy for caluculating amounts of peastone and mulch, but I've not had to use it more than a couple times. Usually, the spaces I'm working are conventional rectangles, squares, circles. Dumb luck, most probably!

Areas of organic shapes are easily measured by baseline dimensioning and then averaging. That is running out a tape and taking perpendicular measurements at equal intervals from that baseline. You add up all of the lateral dimensions and divide by the number of those dimensions. Then multiply by the length of the baseline. It is accurate and simple.

When rougher esitmates need to be made quickly on not too wavy an area, you can measure the length of the area and then eyeball where the average width is perpendicular to that length and measure that width. Then multiply by the baseline and you know the approximate area.

Ichabod,
The GPS should give good results for timber areas over 200 acres. For smaller tracts the percentage of error may be more than you would want.

cady, the diagrams for the tables are not posted here. I'm working an example problem from the Paul Smith College web site. Check the link at the bottom of this post to go there.

laag gives us another way to estimate area. As they say .."More than one way to skin a cat." We might want to look at how this works.

Below is a graphic of an "organic" shape on a grid with a base line shown. Perhaps laag or anyone else could suggest an interval distance to use on this model. Hopefully the grid is clear enough give the base line to shape edge distances. Anyone care to work it with me? Refer to laag's post above for instructions.

{{gwi:42816}}

Thanks for the link to the example, Pls8xx. That one came through intact.

(Quote)The GPS should give good results for timber areas over 200 acres. For smaller tracts the percentage of error may be more than you would want. (Unquote)

Not necessarily. I just "surveyed" my property using my Garmin Map76. To save time on the calculations I used the metric calculator at the link below and got 2.49 Hectares, which is 6.15 acres. My official plat shows it to be 6.46 acres, or less than 5% error. Might have gotten it closer but I have lost two of my corner pins.

I was simply suggesting a quick easy way to get the coordinates in the field. Been wondering, how do you get them?

Ichabod

Ichabod,

My comments on the precision of GPS for small tracts was based on what is generally acceptable in today's world. If you go to a meat market to buy steak you expect the scale used to measure it to be better than plus or minus 5%. Same thing at the gas pump. Most commercial ventures of any type operate on a final bottom line margin of not much more than 10%. Logic would hold that quantity management of any part of the operation is desired to be less than 1 to 2%.

Put yourself in a job, or even a marrige, where the profit margin is 1% and everyone gets stressed out. It ain't fun. I hate accounting, being a bean counter is one small step up from being a ditch digger. They are a neccessary evil to maintain a profit, but they tend to be penny wise and pound foolish. You can't let 'em run the show.

I would expect GPS to get 1% precision on tracts greater than 200 acres. Use of two GPS units and a radio where both units are read at the same time, and coordinates are taken based on the difference in position, should give 1% precision on tracts greater than 40 acres. The expensive differential GPS units that surveyors use will get high precision on tracts as small as 2000 sq ft.

You ask how I would obtain coordinates. It would depend on the cimcumstance. For areas under about 30,000 sq ft I would make taped measurements from two perpendicular base lines. For blocky shaped tracts up to 10 acres I would probably do angle and distance measurements from a central location. Tracts larger than 10 acres and smaller than 100 acres need a survey crew. Above that and GPS takes over.

When making complex calculations one should always seek to employ some element of redundancy as a check for gross errors and blunders. Look back at the last graphic I posted. If the grid units are in feet it is easy to see that the approximate dimensions are about 180 by 95, or an area of 17,100 sq ft. Any calculation of area that is way off this amount will be wrong.

When I invited others to estimate the area of the shape, I failed to tell you that it is composed of standard geometrical shapes, segments of pure circular curves connected with straight lines. As such, the shape can be subdivided into its components and the true and exact area can be determined. I did this and the true area of the shape is 16,303.5

Then I made estimates of the area by the baseline average method suggested by laag. For the first run I used an interval of 20 feet. The 20 data points used, left and right, were:
(0,0) (0,0) (20,56) (20,-30) (40,59) (40,-49) (60,56) (60,-48) (80,52) (80,-43), (100,53) (100,-43) (120,55) (120,-50) (140,50) (140,-55) (160,41) (160,-48) (180,24) (180,0).

The area derived was 16,240 or 99.6% of the true value.

For a second run I chose an interval of 30 feet.
(0,0) (0,0) (30,56) (30,-40) (60,56) (60,-48) (90,51) (90,-40) (120,55) (120,-50) (150,46) (150,-54) (180,24) (180,0)

The result was 15,606 or 95.7% or the true value.

These two calculations demonstrate what you can expect from the baseline average method of estimating area. The precision is dependant on the shape, the interval selected and how the intervals fit the shape. It can give good results but errors of plus or minus 5% are not uncommon.

Are you wondering what I would have done in the field? In the graphic below I show how I would have walked around the shape and set points such that the straight line between them was a good representation of the boundary.

{{gwi:42818}}

Calculation is done as I outlined at the start of this thread.
The data points are:
(0,0) (15,50) (48,62) (85,50) (115,55), (150,48) (180,25) (180,0) (167.5,-42.5) (150,-55) (135,-55) (90,-40) (45,-52) (19,-30)

The result is 16,270 or 99.8% of the true value. This method always yields high precision values, the measurements required are not more than needed for other methods, and the math is just not that complicated.

I am trying to figure out the square footage of our property. The measurements are: back= 180.58, left side= 134.71, right side= 163.91, front= 51.21. I know this is an irregular polygon, but I do not know how to calculate this. Please help.

The calculation cannot be done from the info given.

I am going to guess that the measurements were taken from some type of survey plat. Look and see if directions of the lines are shown (such as N23 47' W), or maybe angles at the line intersections are shown.

The calculation can be done with the addition on one angle or the direction of two connecting lines.

I use SketchAndCalc's web app (although I understand there's an Android App now) and save a bunch of time and head scratching. Just draw the area and it'll spit out square area and perimeter results.

Here's the link to the Area Calculator should you need it.