Just looking for the best price on a 3 ft vinyl fence for about 240 ft of fencing. If anyone has had any good experiences with certain companies please let me know.|||Try your local Home Depot or Lowes they sell the fencing and wil recommend a contrator to install it for you! Ask for references and ask where they have done other jobs to get a feel for what their work is like!
Good Luck
Thursday, November 24, 2011
Can you help with my garden fencing project?
I am replacing my garden fencing with some friends this weekend. It will be shiplap wooden panels with concrete gravel boards and concrete posts.Overall I will be installing 10 6'0" panels.
The garden slopes away from the house and I will have to step down the panels to the bottom end of the garden. Also I will have to start at the top of the garden, where there already is good substantial fencing.
Does anyone have a simplictic way of calculating the step-down I will need for each panel?|||you only need to know one thing
how much is the drop in the land from top to bottom of slop
after you have placed the first one in place at the top the slope.
then measure the drop and if you are placing 9 more panels then divide the measurement by 9 and lower each panel by that amount|||Put the first one in level, and drop each one by 4". This should be able to take the slope.|||First, measure a diameter|||set post at top. fit temporary post at bottom, tie string line over top and top of bottom post, from the string line you will be able to measure down to ground level at the post spaces. there could be problems if the ground is on an incline but also curved? the actual panels would be in steps. you could plan it on paper first.|||Without seeing it its hard to say
If the slopes really steep then you could step it down with the gravel boards (so the bottom of one end is just under the ground and the other end is above ground level) so that each section steps it down by the depth of the gravel board|||As above re the calculations but you also need to remember that if you are using concrete posts that you should use close board sections. Shiplap wooden fence panels are not designed to go with concrete posts and will blow out in heavy wind.|||just build the panel in place put the post in string from top to buttom and fill in the rest it will come out perfect.
The garden slopes away from the house and I will have to step down the panels to the bottom end of the garden. Also I will have to start at the top of the garden, where there already is good substantial fencing.
Does anyone have a simplictic way of calculating the step-down I will need for each panel?|||you only need to know one thing
how much is the drop in the land from top to bottom of slop
after you have placed the first one in place at the top the slope.
then measure the drop and if you are placing 9 more panels then divide the measurement by 9 and lower each panel by that amount|||Put the first one in level, and drop each one by 4". This should be able to take the slope.|||First, measure a diameter|||set post at top. fit temporary post at bottom, tie string line over top and top of bottom post, from the string line you will be able to measure down to ground level at the post spaces. there could be problems if the ground is on an incline but also curved? the actual panels would be in steps. you could plan it on paper first.|||Without seeing it its hard to say
If the slopes really steep then you could step it down with the gravel boards (so the bottom of one end is just under the ground and the other end is above ground level) so that each section steps it down by the depth of the gravel board|||As above re the calculations but you also need to remember that if you are using concrete posts that you should use close board sections. Shiplap wooden fence panels are not designed to go with concrete posts and will blow out in heavy wind.|||just build the panel in place put the post in string from top to buttom and fill in the rest it will come out perfect.
How can I continue fencing in college?
I am going to college in Newport, Rhode Island and the school does not have a fencing team or club. I want to continue fencing because I fenced through high school and middle school.
What can i do to continue fencing?|||Go to the USFA web site and you will find a listing of fencing clubs all over the country.
Rob
What can i do to continue fencing?|||Go to the USFA web site and you will find a listing of fencing clubs all over the country.
Rob
What would the maximum area of a rectangular field that is enclosed by 2400 m of fencing?
I have to determine the maximum area of a rectangular field that can be eclosed by 2400 m of fencing. I have no idea what I'm supposed to do. Someone please help me! It's due on Monday.|||Let sides of rectangular are a and b, then
2(a+b) =2400
a+b =1200
a =1200-b
area=A = a*b = (1200-b)*b =1200b -b^2
A' =1200-2b
for maximum area A' =0
1200-2b =0
2b =1200
b =600 m
a = 1200-600=600 m
maximum area = 600*600 =360000 square metres.|||I'll give you 2 ways.
1) The maximum area of a rectangle occurs when that rectangle is a square.
In your problem, that means each side is 600m, which means the area is 360000m虏.
2) Form an equation.
P = 2(l + w) = 2400
----%26gt; l + w = 1200
----%26gt; l = 1200 - w
A = w(1200 - w)
A = -w虏 + 1200w
axis of symmetry --%26gt; x = -b/2a = 600
Use this to find y-coordinate of vertex.
(600, 360000)
The maximum occurs when(x-coordinate) x = 600m and is(y-coordinate) 360000m虏.|||The field is a rectangle, which means it has length and width. Say W is the width and L is the length.
The perimeter of the rectangle is 2W + 2L. The area is WL.
2W + 2L = 2400m
2L = 2400m - 2W
L = 1200m - W
So now you can replace L in the area with that expression.
A = WL = W(1200m - W)
A = -W^2 + (1200m)W
This equation is a parabola, pointed downward; think of A as y and W as x. To find the center point of a parabola in the form of y = ax^2 + bx + c, you take x = -b/(2a). Here a=-1 and b=1200m, so W = -1200m / -2 = 600m. That means L = 1200m - W = 600m, so the rectangle is a square. The area of this square is (600m)(600m) = 360,000 m^2.|||Suppose one side is x; then the other is 1200-x; So the area is f(x)=x*(1200-x)=1200x-x^2. This is a parabola.
It's maximum is there where f'(x)=0; f'(x)=1200-2x. f'(x)=0 when x=600. So it is a square. And the surface is 360.000 m^2.|||a square has the largest area for a given perimeter. So if you have 2400m/ 4sides = 600m sides. Then, multiply the side by the side to get area- 600*600=360,000m^2.|||perimeter = 2400 m
perimeter = 2x+(2400 m-2x)
Area=x(1200 m-x)
f(x)=x(1200 m -x)
max{f(x) = x (1200 m -x)} = 360000 m虏 at x = 600 m|||A rectangle has maximum area for a given perimeter when all 4 sides are congruent. Or say when it's a square
so the max area for 2400 m fencing is
(2400/4)^2=360000 sq m
2(a+b) =2400
a+b =1200
a =1200-b
area=A = a*b = (1200-b)*b =1200b -b^2
A' =1200-2b
for maximum area A' =0
1200-2b =0
2b =1200
b =600 m
a = 1200-600=600 m
maximum area = 600*600 =360000 square metres.|||I'll give you 2 ways.
1) The maximum area of a rectangle occurs when that rectangle is a square.
In your problem, that means each side is 600m, which means the area is 360000m虏.
2) Form an equation.
P = 2(l + w) = 2400
----%26gt; l + w = 1200
----%26gt; l = 1200 - w
A = w(1200 - w)
A = -w虏 + 1200w
axis of symmetry --%26gt; x = -b/2a = 600
Use this to find y-coordinate of vertex.
(600, 360000)
The maximum occurs when(x-coordinate) x = 600m and is(y-coordinate) 360000m虏.|||The field is a rectangle, which means it has length and width. Say W is the width and L is the length.
The perimeter of the rectangle is 2W + 2L. The area is WL.
2W + 2L = 2400m
2L = 2400m - 2W
L = 1200m - W
So now you can replace L in the area with that expression.
A = WL = W(1200m - W)
A = -W^2 + (1200m)W
This equation is a parabola, pointed downward; think of A as y and W as x. To find the center point of a parabola in the form of y = ax^2 + bx + c, you take x = -b/(2a). Here a=-1 and b=1200m, so W = -1200m / -2 = 600m. That means L = 1200m - W = 600m, so the rectangle is a square. The area of this square is (600m)(600m) = 360,000 m^2.|||Suppose one side is x; then the other is 1200-x; So the area is f(x)=x*(1200-x)=1200x-x^2. This is a parabola.
It's maximum is there where f'(x)=0; f'(x)=1200-2x. f'(x)=0 when x=600. So it is a square. And the surface is 360.000 m^2.|||a square has the largest area for a given perimeter. So if you have 2400m/ 4sides = 600m sides. Then, multiply the side by the side to get area- 600*600=360,000m^2.|||perimeter = 2400 m
perimeter = 2x+(2400 m-2x)
Area=x(1200 m-x)
f(x)=x(1200 m -x)
max{f(x) = x (1200 m -x)} = 360000 m虏 at x = 600 m|||A rectangle has maximum area for a given perimeter when all 4 sides are congruent. Or say when it's a square
so the max area for 2400 m fencing is
(2400/4)^2=360000 sq m
What size fencing would I need for one horse?
I'm getting one horse, and I was wondering how big the fenced in area would have to be and an estimated price on fencing. I'm getting the wood fencing so I'd like an estimated price please.|||The bigger the better. You need to ask your local lumber yard and see what they have. The prices are so different depending on where you live. You can also ask a Home Depot or Lowe's, if you have anything like that nearby.
Find out what you can afford to pay and then find out what size you can make for your horse. A basic rule of thumb is one acre, one horse.
Good luck, sorry I couldn't give more information.|||only your lumber yard can answer this question, the price of lumber varies to much from state to state, in Indiana a horse needs one acre by law, check out you rules in your county.
Find out what you can afford to pay and then find out what size you can make for your horse. A basic rule of thumb is one acre, one horse.
Good luck, sorry I couldn't give more information.|||only your lumber yard can answer this question, the price of lumber varies to much from state to state, in Indiana a horse needs one acre by law, check out you rules in your county.
What is the largest possible area of 5 pens with 600ft of fencing?
A rancher wants to enclose a rectangular area and then divide it into 5 pens with fencing parallel to one side of the rectangle. He has 600 feet of fencing available to complete the job. What is the largest possible total area of the five pens?
How do you set up a problem like this to find area?|||consider the entire figure:
You have a rectangle with 4 lines going though the middle (to subdivide it into 5 pens). Let x be the side parallel to the middle fences, and let y be the other side. Note that, by the confines of the problem,
2y + 2x + 4x = 600, since you have 600 ft of fencing. This simplifies to:
2y + 6x = 600
y = 300 - 3x
This is useful because now you can write the total area of the rectangle as
A = xy = x(300 - 3x) = 300x - 3x^2
Now, finding the maximum is just a matter of differentiating and finding the critical points:
A' = 300 - 6x = 0
6x = 300
x = 50
Thus, your maximum area occurs when you choose a length x of 50. Your total area is:
A = x(300 - 3x) = 50(150) = 7500 square feet
Thank you for the correction Siress :-)
good luck ranching!|||Picture this:
_____
| | | | | |
| | | | | | y
| | | | | |
---------
x
So:
6y + 2x = 600 (feet of fencing)
y = 100 - x/3
Area:
f(x)=x*y = 100x -(x^2)/3
[this is the easy method, requires you to take a derivative]
f'(x) = 100 - 2*x/3
f'(x) = 0 at max and min, since this is 2ed degree, only has one or the other; since x^2 is negative, it's a maximum.
100 - 2*x/3 = 0
x= 150
y= 100 - 150/3 = 50
Self check:
6(50) + 2(150) ?= 600 ....check.
Then, the maximum area is 150*50 = 7500
EDIT: Ben, you need to make a correction:
"A = x(300 - x) = 50(250) = 12500 square feet"
A = x(300 - 3*x) = 7500|||5 squares, so divide 600 by 5, 120
120 ft per pen
divided by 4, 30x30 pens
30x30=900
900*5=4500sqr feet
How do you set up a problem like this to find area?|||consider the entire figure:
You have a rectangle with 4 lines going though the middle (to subdivide it into 5 pens). Let x be the side parallel to the middle fences, and let y be the other side. Note that, by the confines of the problem,
2y + 2x + 4x = 600, since you have 600 ft of fencing. This simplifies to:
2y + 6x = 600
y = 300 - 3x
This is useful because now you can write the total area of the rectangle as
A = xy = x(300 - 3x) = 300x - 3x^2
Now, finding the maximum is just a matter of differentiating and finding the critical points:
A' = 300 - 6x = 0
6x = 300
x = 50
Thus, your maximum area occurs when you choose a length x of 50. Your total area is:
A = x(300 - 3x) = 50(150) = 7500 square feet
Thank you for the correction Siress :-)
good luck ranching!|||Picture this:
_____
| | | | | |
| | | | | | y
| | | | | |
---------
x
So:
6y + 2x = 600 (feet of fencing)
y = 100 - x/3
Area:
f(x)=x*y = 100x -(x^2)/3
[this is the easy method, requires you to take a derivative]
f'(x) = 100 - 2*x/3
f'(x) = 0 at max and min, since this is 2ed degree, only has one or the other; since x^2 is negative, it's a maximum.
100 - 2*x/3 = 0
x= 150
y= 100 - 150/3 = 50
Self check:
6(50) + 2(150) ?= 600 ....check.
Then, the maximum area is 150*50 = 7500
EDIT: Ben, you need to make a correction:
"A = x(300 - x) = 50(250) = 12500 square feet"
A = x(300 - 3*x) = 7500|||5 squares, so divide 600 by 5, 120
120 ft per pen
divided by 4, 30x30 pens
30x30=900
900*5=4500sqr feet
What are the basic Absolutely basic needs for fencing how do I get started?
My daughter is only 12 and she has her mind absoulutely ffixed on fencing!|||Well first you need to find her a place to fence at. I suggest going there and talking to the instructors about what they can provide for her (in case she decides she doesn't like it since equipment is costly). You can start buying her more of her own equipment as she shows steady commitment unless you have the money to spare.
But simply for competition you'll need:
fencing shoes (she can use tennis shoes to start out)
long socks (softball or soccer socks are fine)
spandex shorts (optional but I find the knickers much more comfortable with spandex)
knickers
chest protector (since she's a female this is required for her to fence)
underarm protector
fencing jacket
mask (the type depends on what weapon she fences)
body cords (needed to fence in competitions and at most clubs)
electric weapon (type depends on which she chooses and it is best to have someone who knows fencing walk you through the first purchase so that the weapon suits her)
practice weapon (optional if she always fences electronically)
head cord (if she decides to fence sabre)
glove (must be a fencing glove)|||First, look for a fencing studio in your area, most of these places will charge you a low monthly fee, and will let you use their equipment for no additional cost. Another reason you should do this is because you probably don't know what weapon your daughter will be using. (epee, foil, or sabre) It would be a waste of your money to buy her a foil, then have her coach tell you she'd be better at epee.
After your daughter has practiced for a while, you should buy her a weapon, mask, jacket, chest protector, weapon, and glove. I recommend buying a front zip jacket, and buying an electric weapon, and (if she's sabre) an electric mask. This way, when she does start competing, you won't need to replace her non electric equipment with electric equipment, saving you money.
When your daughter is ready to compete, make sure she has two working weapons, (they have a tendency to break) two body chords, knickers, socks, shoes, a mask, a jacket(electric if she's sabre), a lame(if she's foil), and a glove.
If you have any other questions, email me at wiccan01930@gmail.com|||Most fencing salles have equipment available for use either very inexpensively or free. www.bladefencing.com and www.absolutefencing.com both have "starter kits" that include a jacket, mask, blade, and if she desires to compete soon, an electric weapon and body cord. They tend to run between $150 and $300.
But simply for competition you'll need:
fencing shoes (she can use tennis shoes to start out)
long socks (softball or soccer socks are fine)
spandex shorts (optional but I find the knickers much more comfortable with spandex)
knickers
chest protector (since she's a female this is required for her to fence)
underarm protector
fencing jacket
mask (the type depends on what weapon she fences)
body cords (needed to fence in competitions and at most clubs)
electric weapon (type depends on which she chooses and it is best to have someone who knows fencing walk you through the first purchase so that the weapon suits her)
practice weapon (optional if she always fences electronically)
head cord (if she decides to fence sabre)
glove (must be a fencing glove)|||First, look for a fencing studio in your area, most of these places will charge you a low monthly fee, and will let you use their equipment for no additional cost. Another reason you should do this is because you probably don't know what weapon your daughter will be using. (epee, foil, or sabre) It would be a waste of your money to buy her a foil, then have her coach tell you she'd be better at epee.
After your daughter has practiced for a while, you should buy her a weapon, mask, jacket, chest protector, weapon, and glove. I recommend buying a front zip jacket, and buying an electric weapon, and (if she's sabre) an electric mask. This way, when she does start competing, you won't need to replace her non electric equipment with electric equipment, saving you money.
When your daughter is ready to compete, make sure she has two working weapons, (they have a tendency to break) two body chords, knickers, socks, shoes, a mask, a jacket(electric if she's sabre), a lame(if she's foil), and a glove.
If you have any other questions, email me at wiccan01930@gmail.com|||Most fencing salles have equipment available for use either very inexpensively or free. www.bladefencing.com and www.absolutefencing.com both have "starter kits" that include a jacket, mask, blade, and if she desires to compete soon, an electric weapon and body cord. They tend to run between $150 and $300.
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