depends upon stage of production and lbs of dry matter yield.
lactating ewes eat approx 8 lbs dm per day, dry ewes eat approx 4.5 lbs per day. so assuming you are grazing in the fall, these are dry ewes, so that is 200 x 4.5 = 900 lbs dm per day. 810 rolls fence in approx 3.5 acres (?) so if that land has approx 2100 lbs dm per acre and you want to leave behind 900 lbs per acre (if pasture) then that means you have 1200 lbs per acre to consume or a total of 4200 lbs for 3.5 acres. 4200/900 = 4.6 days
I am not familiar with lbs dm intake for stalks or lbs per acre.
Janet
Grazing Corn Stalks

 Old Hand
 Posts: 6037
 Joined: Fri Mar 24, 2006 12:26 pm
 Location: East Central Minnesota
 Contact:

 Chief Shepherd
 Posts: 5903
 Joined: Thu Mar 23, 2006 9:27 pm
 Location: Westmoreland, NH
 Contact:
My only quibble with those figures is that with eight rolls of net the best you're going to get is a little shy of 2.5 acres. If the eight rolls are set up in a square, you've got 328 feet to a side, or 107,584 square feet. Divided by 43560 equals 2.47 acres.
With 10 nets you can get that up to 3.8 acres.
If your paddocks are rectangular or irregular, the enclosed area will go down.
(Actually, the most efficient use of fence would be a perfect circle. Bonus points to anyone who figures out the area of a circle enclosed by 10 rolls of electronet.)
With 10 nets you can get that up to 3.8 acres.
If your paddocks are rectangular or irregular, the enclosed area will go down.
(Actually, the most efficient use of fence would be a perfect circle. Bonus points to anyone who figures out the area of a circle enclosed by 10 rolls of electronet.)
Bill Fosher
Westmoreland, NH
Westmoreland, NH

 Old Hand
 Posts: 6037
 Joined: Fri Mar 24, 2006 12:26 pm
 Location: East Central Minnesota
 Contact:
Hi Bill,
I'll trust your figuring. I know I one time figured it out to be close to 3.5 acres (rectangle) but could not remember if it was 8 or 10 rolls I used in the figure.
Snowing to beat the band here. County says they are going to shut down the plows after dark.
Janet
I'll trust your figuring. I know I one time figured it out to be close to 3.5 acres (rectangle) but could not remember if it was 8 or 10 rolls I used in the figure.
Snowing to beat the band here. County says they are going to shut down the plows after dark.
Janet
Janet McNally
Tamarack Prolific and Ile de France crosses
Minnesota
Tamarack Prolific and Ile de France crosses
Minnesota

 Old Hand
 Posts: 394
 Joined: Fri Mar 24, 2006 11:15 pm
 Location: SW WI USA
 Contact:
Other than the pi number use feet.
Area of a circle is pi, (or 3.1416 ) x radius squared.
To find the distance around the outside of a circle, multiply pi,(3.1416) x diameter.
And yes the biggest capture area for any amount of length is in a circle.
I dont know how long 10 lengths of net are but you can figure with these formulas and then divide by 43560 to get acres.
To get the diameter if you know the length, (10 rolls of net) divide by 3.1416 and you have the diameter.
Area of a circle is pi, (or 3.1416 ) x radius squared.
To find the distance around the outside of a circle, multiply pi,(3.1416) x diameter.
And yes the biggest capture area for any amount of length is in a circle.
I dont know how long 10 lengths of net are but you can figure with these formulas and then divide by 43560 to get acres.
To get the diameter if you know the length, (10 rolls of net) divide by 3.1416 and you have the diameter.

 Chief Shepherd
 Posts: 5903
 Joined: Thu Mar 23, 2006 9:27 pm
 Location: Westmoreland, NH
 Contact:
Noticed nobody answer the question, when browsing around.
Since we know our perimeter of the square is 1312 (328 * 4), we can then find the diameter of the transformed circle, 417.83 (C/Pi = D). Since area is A=Pi/4(D to the second) hence 3.14/4 * (174,581.91) = 137,046.80 as to the acre you get 3.15. I am using PI/4 here because it explains better what you are trying to do.
If I need to do a quick ballpark figure I normally add 20% to the square area, since PI/4 represents the ratio circumscribed square. Since you had 2.5 acres I to 20 percent of that (.5 acres) my ballpark would be 3 acres.
FYI the next a time a youngster ever ask why do I need Trig, this may be one example. It also comes in handy when you are doing home improvements (figuring the area of attic, building stairs that need to be at 33% incline, ect).
Bill, do I get an A for the day, assuming I am right.
John
Since we know our perimeter of the square is 1312 (328 * 4), we can then find the diameter of the transformed circle, 417.83 (C/Pi = D). Since area is A=Pi/4(D to the second) hence 3.14/4 * (174,581.91) = 137,046.80 as to the acre you get 3.15. I am using PI/4 here because it explains better what you are trying to do.
If I need to do a quick ballpark figure I normally add 20% to the square area, since PI/4 represents the ratio circumscribed square. Since you had 2.5 acres I to 20 percent of that (.5 acres) my ballpark would be 3 acres.
FYI the next a time a youngster ever ask why do I need Trig, this may be one example. It also comes in handy when you are doing home improvements (figuring the area of attic, building stairs that need to be at 33% incline, ect).
Bill, do I get an A for the day, assuming I am right.
John

 Old Hand
 Posts: 723
 Joined: Tue Jan 16, 2007 11:05 am
 Location: Corvallis Oregon
John,
Another math question for you.
One time I was offered a 90 acre circle of rape and sudangrass for grazing. One stipulation on the use of the forage was that I could not build the perimeter fence in such a way that any of the wire or posts were in the road surrounding the circle. I built the fence using 2 strands of 14 gauge smooth wire Tposts for corners and 1/4 inch fiberglass posts. I pride my self on building a good tight fence and made the assumtion that the fiberglass posts were only strong enough to cheat the 2 wire fence a foot and a half out of a straight line when encircling the field. How many tposts did it take to encircle the field and keep the wire at the edge of the road and out of the field?
Tom Nichols
corvallis, Orregon
Another math question for you.
One time I was offered a 90 acre circle of rape and sudangrass for grazing. One stipulation on the use of the forage was that I could not build the perimeter fence in such a way that any of the wire or posts were in the road surrounding the circle. I built the fence using 2 strands of 14 gauge smooth wire Tposts for corners and 1/4 inch fiberglass posts. I pride my self on building a good tight fence and made the assumtion that the fiberglass posts were only strong enough to cheat the 2 wire fence a foot and a half out of a straight line when encircling the field. How many tposts did it take to encircle the field and keep the wire at the edge of the road and out of the field?
Tom Nichols
corvallis, Orregon
I would initially say zero because you do not have corners but assuming I know what you meant.
I know my math somewhat pretty good, life of a programmer I guess. I do not know how far you can go before you would need a TPost to keep the circle from collapsing in. I for one would guess you would not being making a circle but rather some sort of polygon because you have stress points; then again I am new at building fences. I will not go into COS and SIN because this is a Sheep Board and not a Math Board.
To answer your question I would need to know have far you are putting your TPost, if it is every 25 feet then you would need about 278, if it were 10 feet then 701 TPost. The circumference of the circle is 7017.15, you divided that by 10 you get 702 (if you round to a whole post), and then minus one because you come back to the beginning.
A=PI/4 * D^2, 90 acres is 3,920,400 = .785 * (D^2), 4,994,140.13 = D^2, D = Square Root (4,994,140.13) or 2234.76. Now that we know our diameter we can find the circumference by multiplying by PI giving us 7017.15. To make sure your right you can validate by the Trig formula of 90(43560) = .785(2234.762)^2 or by the Geometry formula of 90(43560) = (2234.76/2)^2 * 3.14.
Little rusty but then again... Hopefully I typed it all correctly, now time to learn about Sheep. I am not doing your homework am I
John
I know my math somewhat pretty good, life of a programmer I guess. I do not know how far you can go before you would need a TPost to keep the circle from collapsing in. I for one would guess you would not being making a circle but rather some sort of polygon because you have stress points; then again I am new at building fences. I will not go into COS and SIN because this is a Sheep Board and not a Math Board.
To answer your question I would need to know have far you are putting your TPost, if it is every 25 feet then you would need about 278, if it were 10 feet then 701 TPost. The circumference of the circle is 7017.15, you divided that by 10 you get 702 (if you round to a whole post), and then minus one because you come back to the beginning.
A=PI/4 * D^2, 90 acres is 3,920,400 = .785 * (D^2), 4,994,140.13 = D^2, D = Square Root (4,994,140.13) or 2234.76. Now that we know our diameter we can find the circumference by multiplying by PI giving us 7017.15. To make sure your right you can validate by the Trig formula of 90(43560) = .785(2234.762)^2 or by the Geometry formula of 90(43560) = (2234.76/2)^2 * 3.14.
Little rusty but then again... Hopefully I typed it all correctly, now time to learn about Sheep. I am not doing your homework am I
John

 Chief Shepherd
 Posts: 5903
 Joined: Thu Mar 23, 2006 9:27 pm
 Location: Westmoreland, NH
 Contact:
Who is online
Users browsing this forum: No registered users and 1 guest