Number - Plants / Hills per Acre

[Guest]

"What is the question...........???"

A farmer has a crop to plant in an acre. He plans to use 3' 6" rows, 3 plants per hill, hills 3' 6" apart. Determine number of plants.

[Guest]

Are we trying to answer the question or getting a question to fit the answer...?????
This was stated in earlier post.......a rectangular acre can be any size as long as the width x length = 43560
Below is a table showing up to 50 rows 3 ft 6 ins apart and hill spacing 3 ft 6 ins and all have total 10667 plants because it doesnt matter how many as long as width x length is 43560.

It is pasted here as text you will have to count the columns to match the heading as they are not aligned due to pasting.

Rows Acre sqft Width Length Spacing plant/row plants/hill Total plants
1 43560 3.5 12446 3.5 3556 3 10667
2 43560 7.0 6223 3.5 1778 3 10667
3 43560 10.5 4149 3.5 1185 3 10667
4 43560 14.0 3111 3.5 889 3 10667
5 43560 17.5 2489 3.5 711 3 10667
6 43560 21.0 2074 3.5 593 3 10667
7 43560 24.5 1778 3.5 508 3 10667
8 43560 28.0 1556 3.5 444 3 10667
9 43560 31.5 1383 3.5 395 3 10667
10 43560 35.0 1245 3.5 356 3 10667
11 43560 38.5 1131 3.5 323 3 10667
12 43560 42.0 1037 3.5 296 3 10667
13 43560 45.5 957 3.5 274 3 10667
14 43560 49.0 889 3.5 254 3 10667
15 43560 52.5 830 3.5 237 3 10667
16 43560 56.0 778 3.5 222 3 10667
17 43560 59.5 732 3.5 209 3 10667
18 43560 63.0 691 3.5 198 3 10667
19 43560 66.5 655 3.5 187 3 10667
20 43560 70.0 622 3.5 178 3 10667
21 43560 73.5 593 3.5 169 3 10667
22 43560 77.0 566 3.5 162 3 10667
23 43560 80.5 541 3.5 155 3 10667
24 43560 84.0 519 3.5 148 3 10667
25 43560 87.5 498 3.5 142 3 10667
26 43560 91.0 479 3.5 137 3 10667
27 43560 94.5 461 3.5 132 3 10667
28 43560 98.0 444 3.5 127 3 10667
29 43560 101.5 429 3.5 123 3 10667
30 43560 105.0 415 3.5 119 3 10667
31 43560 108.5 401 3.5 115 3 10667
32 43560 112.0 389 3.5 111 3 10667
33 43560 115.5 377 3.5 108 3 10667
34 43560 119.0 366 3.5 105 3 10667
35 43560 122.5 356 3.5 102 3 10667
36 43560 126.0 346 3.5 99 3 10667
37 43560 129.5 336 3.5 96 3 10667
38 43560 133.0 328 3.5 94 3 10667
39 43560 136.5 319 3.5 91 3 10667
40 43560 140.0 311 3.5 89 3 10667
41 43560 143.5 304 3.5 87 3 10667
42 43560 147.0 296 3.5 85 3 10667
43 43560 150.5 289 3.5 83 3 10667
44 43560 154.0 283 3.5 81 3 10667
45 43560 157.5 277 3.5 79 3 10667
46 43560 161.0 271 3.5 77 3 10667
47 43560 164.5 265 3.5 76 3 10667
48 43560 168.0 259 3.5 74 3 10667
49 43560 171.5 254 3.5 73 3 10667
50 43560 175.0 249 3.5 71 3 10667
51 43560 178.5 244 3.5 70 3 10667

[Guest]

Thanks.

"Are we trying to answer the question or getting a question to fit the answer...?????" Trying to answer the question.

The answer was with the other statement I discovered.

Was the data in the table calculated with a spreadsheet ? Just curious.

[Guest]

Well, then you would need to be more specific with the question.......
"Are we trying to answer the question or getting a question to fit the answer...?????" Trying to answer the question.

In the question you state the rows are 3 ft 6 ins apart and the plants are 3 ft 6 ins apart with no constraints or specified shape or size for the acre of land. As we can see they can be planted at the corners of a 3 ft 6 ins square, that will put them in rows 3 ft 6ins apart and the plants 3ft 6 ins apart on each row. But that will not give the answer as in the last post.
The table in the last post if you had tried to calculate any row of it, assumes the plants are planted in the middle of a 3 ft 6ins row and in the middle of 3 ft 6ins squares to keep the plants 3 ft 6ins apart. So the plants are still in rows 3 ft 6 ins apart and the plants are 3 ft 6 ins apart, but in the middle of squares. So that means at the edges the plants are half the width of a square in from the edge, which is a totally different scenario from what your question was implying in earlier posts. So that is why the answer is different for each case we are taking. So in actual fact you are trying to "conjure" up a question to suit a particular answer.
And also the rectangular area of an acre can be any size as long as the width x length = 43560 it does not matter the size of it for the last answer you gave.

Regards the spreadsheet it is still calculated by the formula and multiplying and division as before the spreadsheet on keeps track of the repetative calculations for each scenario. I can post a table of 10000 calculations if you want.....

[Guest]

Thanks.

I was not trying to "conjure" up a question as you stated. The numbers stated in the posted question are correct. The shape of the acre was not mentioned, only an acre.

[Guest]

Well then why post an answer without showing the working out.....

Post your solution so that I may see your understanding of the problem and how you work it out.......

When I post a solution i describe what I am doing and how I arrive at the answer.....It may not be the scenario you perceive the problem to be....

I still do not know what a "hill" is but I presume it is some sort of mound or raised soil with plants planted on top......Like square pyramids with a square base. And they are placed on a 3ft 6 ins grid equivalent to rows 3 ft 6 ins apart and plant spacing of 3ft 6 ins........

Below is another example calc. for a rectangular or square plot .............

Again with rows 3 ft 6 ins apart.....Rectangular or Square plot..........
and plant hills 3 ft 6 ins apart on each row.....plant 3 plants in the middle of a hill.....

R = number of rows......3ft 6 ins between each row

Width = 3.5 x R = 3.5R
Length = 43560 / 3.5R

Area = width x Length = 3.5R x 43560 / 3.5R

For a square..... Length = Width..... 3.5R = 43560 / 3.5R

Then.......... 12.25 R^2 = 43560

R = sqRoot (43560 / 12.25) = 59.63 rows ....So plant 59 rows

There will also be 59 plant hills in each row and 3 plants at each hill

Total plants = 3 x 59 x 59 = 10443 plants.

This means there will be a border around the outside of the plot
So for the width will be ..... 0.63 x 3.5 = 2.2 feet
If we add this to the length side we get now 4.4 feet and room for another row of 59 plant hills

So that gives total plants = 3 x 59 x 60 = 10620 plants
And that now only leaves a border of 4.4 - 3.5 = 0.9 feet along one side of the plot.

It is not now square it is 59 rows x 60 plant hills in each row
OR 60 rows with 50 plant hills in each row.

Area planted is 59 x 3.5 x 60 x 3.5 = 43365 sq feet
plus unplanted edge 59 x 3.5 x 0.9 = 185.9 sq feet
which is near enough an acre except for roundings in the calcs.............

[Guest]

"Well then why post an answer without showing the working out....." The calculations were not shown - only the number of plants.

"Post your solution so that I may see your understanding of the problem and how you work it out......." I didn't know how.

"I still do not know what a "hill" is but I presume it is some sort of mound or raised soil with plants planted on top" You are correct - When seeds are covered with soil or plants are surrounded with soil, the soil is in a mound or raised. That is where the "hill" comes from.

One more related question:

The statement also says in an acre there are 10,677 plants with each single plant placed 14 inches apart.


14 inches = 1.1666 feet

208 / 1.1666 = 178.29 = 178 plants per row

How do I proceed ? Thanks.

[Guest]

The statement also says in an acre there are 10,677 plants with each single plant placed 14 inches apart.

This is beyond comprehension and belief...........
You must not have read any of the previous posts OR learned anything from the previous posts.

You have now come up with another scenario for planting within an acre and you don't give any information about the size, shape or layout of either the plot of land or the planting sequence of the plants. So they could be planted in a circle or any other shape.

for 1 acre area = 43560 sq ft = 6272640 sq ins.

For a 1 acre square sides are 2504 inches

14 inches apart = 2504 / 14 ........

Gives 179 rows of 179 in each row = 32041 plants.............
..........................

for 1 acre circle diameter = sqroot(4 x area / pi) = 2826 inches = 1414 ins approx radius

At 14 inches apart there will be 101 concentric circles.....

There will be (pi x 28/14) = 2 x pi = 6.28 plants less in each successive circular row....arithmetic series.

So find the sum of 101 terms from 1 to 101 having a common difference of 6.28

Sum = (101/2)x(2x6 + 6.28(101-1))

Sum = (101/2)x(640) = 32320 plants plus 1 in centre = 32321 plants
........................................

[Guest]

I am sorry I posted the questions, wasted your time, and was a bother.

[Guest]

If it is a real question, then post the full question with all the detail

Also post your solution in detail, logically going through each step

Then we may be able to compare our understanding of the problem and work out a solution

[Guest]

"If it is a real question, then post the full question with all the detail" It is a real question - There are 10,677 plants in an acre, planted in 3' 6" rows, 3 plants per hill with hills 3' 6" apart, or 1 plant every 14 inches. The statement does not give any description of the acre. I read the information in an agricultural pamphlet.

"Also post your solution in detail, logically going through each step" I don't have a solution - I didn't know how to solve.

Since "This is beyond comprehension and belief..........." never mind.

[Guest]

Again, I apologize for wasting your time.

[Guest]

My apology again for posting the questions.

[Guest]

Well then since you give an answer and show no working out, we will start with the answer and see can we fit them in an acre planting as specified.

There are 10,677 plants in an acre, planted in 3' 6" rows, 3 plants per hill with hills 3' 6" apart, or 1 plant every 14 inches.
So the Rows are 3 ft 6 ins apart = 42 inches apart
Hills are 3 ft 6 ins apart, if that means 3 ft 6 ins between centres = 42 inches apart between centres and all hills in rows...
3 plants per Hill, if evenly spaced on the hill will be 42 / 3 = 14 inches between each plant in a row along the middle of the hill. Actually 1 plant at the centre and one either side 14 inches apart leaving 14 (2 x 7) inches between the last plant on one hill and the first on the next hill.
That means there will be rows 3 ft 6 ins apart made up of plants 14 inches apart (between each plant) on hills that are 3ft 6 ins apart OR 42 inches between the centre of each hill. The length of a hill will be 2 x 14 inch plant spaces plus 7 inches between the end plant and hill end.
So equivalent to rows of plants at 14 inch centres throughout and each row will be 42 inches apart.

Now lets keep it mathematicaly speaking....

Let R = number of rows.......
Then for a square acre there will be 3 time more plants along the side of the plot that is parallel with the rows than there will be number of rows. And the number of rows multiplied by the number of plants in a row will equal the total number of plants.

So..... R x 3R = 10677

So...... 3R^2 = 10677

So......... R = sqroot(10677 / 3) = 59.66

So ...... number of plants in a row = 3 x 59.66 = 178.98

Total number of plants alltogether = 3 x 59.66 x 59.66 = 10677
We started with the answer of 10677 so we end with tha answer of 10677

And the plot of land will be 59.66 x 3.5 x 59.66 x 3.5 = 43602 sq feet.

Above is NOT considering roundings for need to plant whole numbers of rows or whole plants in each row.

So if we take whole rows then 59 rows
And whole number plants in each row then 3 x 59 = 177
Then total number of plants is 59 x 177 = 10443 .........same as earlier post
And the plot of land will be = 59 x 3.5 x 59 x 3.5 = 42642 sq feet.

[Guest]

Thanks again. You did not have to respond after I wasted your time.

[Guest]

As you say this is an agricultural question and most acres or plots of land are irregular shapes....and not all nice neat rows.....

So the farmer wants to know the density of planting, that is how many sq.feet a plant occupies OR how many plants per acre etc. etc......
NOT specifically the number of plants in a row exactly and the number of rows exactly......

So we will work it out on a per unit basis (per acre or per sq.foot) and see if we come up with the answer you have. The farmer will know or can measure a small area of the field and use this sample to calculate for the whole area however big, but in this case we will take 1 acre or 43560 sq feet.
And the farmer will want to calculate the number of plants per acre using the sample....we will use the question data as sample figures.....

Thus 3 plants planted on each hill and each hill 3 ft 6 ins apart (42 inches apart) is equivalent to each plant being 14 inches apart OR 1.167 feet apart on the hills and the hills are in rows in one direction and the rows are 3 ft 6 inches apart across the field in the other direction.

(Dist. between plants in the hill rows) x (Dist. between the rows) = (The Area occupied per plant)

1.167 x 3.5 = 4.08 sq feet occupied per plant

And the Plants per Acre = (the sqfeet in an Acre) / (the sqfeet occupied per plant)

So..... Plants per acre = 43560 / 4.08 = 10676 plants per acre..........and we didn't need to know the dimensions of the field.

Farmers always ....... Keep it Simple ........

[Guest]

Thanks again for all the helpful info...... Interesting.

[Guest]

And just another way of looking at it..............
The rows of plants are spaced at 3 ft 6 ins in one direction and the hills are 3 ft 6 ins apart in the other direction and there are 3 plants per hill.
So that is a square grid of 3 ft 6 ins by 3 ft 6 ins.
So the farmer may decide to count the number of plants in a group of adjacent grids and use this to calculate for the whole acre.
So lets take a group of 3 grids by 3 grids......equal to 3 rows and 3 hills on each row............
The area of that is (3 x 3.5) x (3 x 3.5) = 110.25 sq feet.
The number of plants in that area will be 3 rows multiplied by 9 plants in each row = 3 x 9 = 27 plants
So that is 27 plants per 110.25 sqfeet. = (27 / 110.25) = 0.2449 plants per sq.foot.
So the Number of plants per acre = 43560 x 0.2449 = 10667 plants per acre.
Which is the same as the last post result except for roundings due to the number of decimal places used in the calculations.

Farmers ......... Keep it Simple ...........

[Guest]

Again, thanks. How did you think of all the solutions ? I could not do even one.

[Guest]

It's just different ways at looking at the problem depending on the information you start with or what you are trying to calculate from the information.

The first ones were trying to decipher how many plants fitted in a row and how many rows fitted, rounding off to whole number of plants and rows.

The equilateral triangle allows all to be planted equidistant from each other in 6 directions (hexagons). I think it is the most dense for a specific spacing.

The last solutions were based on per unit area and the ratio that up to the total area. This is probably the most practical in real life where the farmer can plant or measure a small area, work out the per unit value and ratio it up to the whole area. We were not considering if whole number plants fitted across the length or width of the large area.