Fence Posts

Depending on the exact shape of the hill, about 4-5 additional posts should do it.

Hello Janissaries,

You have made a question which may have many unexpected answers - this is one of them.

You don't state many variables:

1. The shape or cross-section of the hill in the direction of fence post line - Specify please

2. Whether you are digging a hole, inserting the fence posts, and backfilling - Specify please

3. Whether the fence posts are, if inserted as per (2) above:

• Fixed vertically - Specify please
• Fixed at any particular angle - Specify please
• Fixed at right angles to the terrain - Specify please
• Other - Specify please

4. If the posts are NOT inserted into the ground, which way are they lying:

• Lengthwise along the fence line - Specify please
• Right angles to the fence line - Specify please
• Other - Specify please

5. Post length - Specify please

6. Post material:

• Rigid - Specify please
• Flexible, (like soft rubber), able to follow the ground surface contours) - Specify please
• Other - Specify please

7. Is the intended fence line in a straight line, or are there curves, arcs, angles, or corners - Specify please.

8. There may well be other variables, and perhaps others may think of them, to add into their post reply....

It is, after all, possible that the posts are 20 feet long (suitable for a lion-proof enclosure), and if the 20 foot long posts were placed on the ground (unburied) in line with the intended fence line, and the ends thus 20 feet apart, the number of actual posts required would be far different from what you expect.

If the posts were 40 feet long, (suitable for mad giraffes), for an enclosure which required some 30 feet of the post above ground, and the posts were placed on the ground (unburied) in line with the intended fence line, and the ends thus 20 feet apart, the number of actual posts required would be even less.

You may consider me rather pedantic, but in a question, or Specification, it really is good to leave no loopholes for the lateral thinker to leap through.

If you are involved in Commercial Contracts, the Specification must leave no room for a Contractor, who may later, with a small amount of careful thinking, gain what may escalate to very large extra costs from the paying Client.

Looking forward to hearing back from you, with....

Bin there, done that ...

John.

I read it as a challenge question because the poster specified "you are placing posts", not the poster. I would assume it to be theoretical, so I used perfect circles and straight line.

I assume that the poster was looking for us to solve the problem and show how we solved it, so I did that with a few steps glossed over since they are basic geometry and I did not know how to draw graphics on this web site.

The key question is what does he mean by "apart"? My initial reaction is to support Del (unless there is a place where the hill is so steep that the top of one fence post is lower than the base of the next), but if "apart" is defined by the number of inches along the ground then we need large amounts of intricate detail to solve it.

Reckon:

For flat land: 500yds = 1500ft, 1500/20 = 75, add one on the end = 76 posts.

With the hill, flat bit = 425yds = 1275ft.

Assuming hill approximates to an isosceles triangle:

Sloping bit (all in ft) = √(200² + (75*3)²) = √(40000 + 225²) = √(40000 + 50625) = √90625 = 301 (and a teeny bit).

So total length = 1275+301 = 1576ft.

1576/20 = 78.8 (call it 79), add one on the end = 80.

Need 4 more posts.

I must confess, I've had a drop or two this evening, and I'm tired and I wanna go to bed, so I've prob'ly mucked it up big style. Please check my signature before placing your order.

Here is how to approach the problem. Break it down into chunks. We know the radius of the circle (hill) = (75 yards * 3) / 2 = 112.5 feet.

The whole schematic is pictured as a circle with a chord running horizontally across it. The chord is considered the ground. The radius of the circle is below the chord by 12.5 feet. The circle sits at one end of the total fence line.

Additionally, I will make a triangle that has three sides composed of: Side B = 1/2 of the chord, side A = the length between the radius and the chord, and side C = radius of the circle (from the radius to the end of the chord. I will use that triangle to calculate the subtended angle of the chord. Here we go...

We know the distance from the top of the circle to the chord that intersects it (sag) = 100 feet.

The difference between the sag and the radius is 100 - 112.5 = 12.5 feet. This is the height of a right triangle that has a hypotenuse = radius of the circle = 112.5. Pythagorus gives us the remaining side of that triangle as 111.8 feet (left to the student to figure out), which is 1/2 of the chord through the hill.

Now I can calculate the angle of the triangle between the chord and the radius using arctangent of the rise (12.5 feet) / run (111.8 feet) = 6.38°.

Imagine a triangle from the circle radius to the ends of the chord. I must calculate the angle subtended from the radius to the ends of the chord.

I know that the angle will be 2 * (90° - 6.38°) = 167.2°. Now I can calculate the circumference of the arc.

The circumference of the 112.5 feet radius circle = 2PI * r = 706.9 feet for the complete circle. We only need to know the arc which is 167.2°/360° * 706.9' = 328.4 feet. That is the arc length!

Our total fence line is 500 yards or 1500 feet. Our chord is (2 * 111.8 feet) = 223.6 feet. Subtract that from 1500 - 223.6 = 1276.4 feet. That is the length of the total run minus the hill. Now add the arc length of the hill (328.4 feet) to the length = 1276.4 + 328.4 = 1604.8 feet.

Our extra distance = 1604.8 - 1500 = 104.8 feet.

You need six extra posts because 5 posts get you 100 feet and the extra 4.8 feet require a 6th post. If you want, simply do it in 5 and stretch the distance between some of the posts to make up the difference.

Hello Anonymous Hero,

Top marks for the Maths, the Awards Committee have much pleasure in giving you this fine

But you have allowed for posts of infinite thinness.

The thickness (radius, diameter, length x breadth etc) has yet to be quantified.

I must admit I overlooked that in my earlier Post:

http://cr4.globalspec.com/comment/158192/Re-Fence-Posts

and trusted that a more perspicacious person, such as yourself, would have noted my omission of that variable, and advised that the post cross-sectional shape and area requirement be specified.

For the definitive answer to be correct, we are still waiting for the variables to be clarified.

Looking forward to ....

Good point! However I figured that typically posts are specified center to center and since the diameter wasn't given, then it wasn't important.

However, hats off to you for your diligence. I will bet you are a damn good engineer, because you know the devil is in the details!

It comes down to the interpretation of the "hill 100' high and 75 yards in diameter". I'd taken it as a conical hill 100' high, with base circle diameter = 75yds. Hence my isosceles triangle approximation. Janissaries -need more info (tho' I don't think we need go quite as far as SparkY would like!)

This is not a good question for CR4. A practical approach is needed.

You may need 10 or more additional posts.

First select points with definite gradient changes and plant posts there and at the begin and end.

each section should now be subdivided into equal gaps. The gaps can be stretched on slopes but should be shortened on curves.

The reason is to keep the bottom wire away from the ground and to eliminate gaps.

Happy fencing.

Off topic . Do you plan to use a few stronger posts?

Del - Now that is a good answer!!!!

John.

Good answer, yes, but the fence will look like the hind leg of a cat, I mean lion I mean dog.

Actually it wouldn't it have to be a pretty severe hill, or stupidly widespaced fence posts to make much difference?

With my conical hill approximation, I reckon spacing up the slope (for 20' horizontal pitch) would be about 26' 9" (back of screwed-up envelope, so could be wrong).

I apologize for not posting sooner, I was gone for these last couple weeks.

Del is correct.

It doesn't matter what the shape of the hill is.

You measure the distance as if the ground was flat all the way down. The posts are planted verticle with the horizontal surface. you would need more wire between the posts going up and down the hill.

If you were planting them as perpendicular with immediate surface you would need more posts but that isn't the correct way of doing it.

My Dad asked me this question and I thought it would be a good challenge to present to you guys to work with for fun.

you can challenge us, but as you see how things got side track...I would advice you not to ask us to help.

Yes, I may not be a guru but I've viewed enough to see all the different directions comments go and I am no exception.

I was surprised with the answer when I found out about it and figured people would have fun with it.

So d'ya mean this was a wynd-up? I spent fifteen hours sorting out your dam' fence for nuthin'? Paa.

I figured after a week and a half everybody would be expecting one.

help build it that is.

"It doesn't matter what the shape of the hill is.

You measure the distance as if the ground was flat all the way down."

Sez who?

Program on the Rural Farming (RFD) cable channel.

Answers 3

This reminds me of the surveying lore from the Canadian prairies. Rumour has it that the chief land surveyor was out from Ottawa and found one of his subjects driving along with his buggy and a handkerchief tied to it. When he asked him about it the young surveyor said, "Why I know the wheel's circumference, so by counting the wheel turns, I know the distance travelled & Voilla, it saves chaining." The supervisor said, " but what happens when you go uphill." "Why, I make it up coming down on the other side" was his smart aleck reply. Well, apparently this was standard procedure, and the skewed prairie roads in Western Canada attest to it. No Del, this time you'd have to go right through the hill to come up with the same number of posts.

I guessed an elipse for this hill, and came up with an extra 113 feet. Oh, and my father taught me to fence. He was a perfectionist.

Nope, you just place 'em as if marked equidistant on a map.

I didn't say it was easy, but with a good contour map and a bit of geometry (or GPS) you could work out the 'along the ground' distance to give the correct plan (horizontal) distance. It may be bad, silly, impractical...but it isn't 'wrong'

Side question; How is an acre measured on very uneven ground? If you own one square acre, is this a plane 209x209 or 209x209 surface measure (up and down hills)?

So far nobody has mentioned another obvious point.

Nowhere in the question is it stated the fence line is in a straight line.

I had expected some poster would realize that the fence line can be any shape, providing "The distance on a flat surface is 500 yards from one end of the fence to the other." and "You're placing fence posts 20 feet apart".

Thus the actual fence line may be any shape, zig-zags, arcs, angles, ellipses or circular (with a piece cut out leaving a 500 yard gap between the end posts), corners etc. and still comply with the stated requirement of: "The distance on a flat surface is 500 yards from one end of the fence to the other." and "You're placing fence posts 20 feet apart".

So...........

Are there no surveyors here? This seems like a problem that would yield to some simple, and very basic survey calculations. Just wondering. I'll go see what I can find.

I think it's more a case of 'we can't be arsed'.... hence my smart-arse answer .

Del

Del, what if one or more posts got broken in installation - that could add to the number needed to go over the hill. Tell them to buy extra posts.

Yeh, those fence posts are sooo delicate .

Anyone sensible buys say 5% over to allow for contingencies...or if not, they have a few left from a previous job.
The whole question is bonkers..no one is going to survey it to the exact number of fence posts, any extra can be used to double up at awkward corners, gates etc, or to make giant crossbow bolts...whatever.

Anyhow you would probably round up the order to the nearest pallet load or armfull or.... why am I still writing this stuff?

Del

I have done surveying 45+ years ago.

We do need full information, before any useful calculations may be done.

Come back, Janissaries, with ....

I'd go around one side of the hill if I could. That slope is going to approximate 45^o, so the additional work/time of lugging stuff up and down just isn't worth it. The extra material cost will be off-set by the time saving. Is this a real situation you have Janissaries , or just a 'fun' type calculation exercise ? If it's just for fun I'd let somebody pay me to dynamite the thing out the way if it's made of useful rock. A slope like that doesn't sound much use for anything. Cattle would topple over, and a Combine Harvester would struggle with the gradient.

It could be that going (a)round will cause the fence to traverse a road/right-of-way. Adding a gate there, along with additional, heavier posts and tension posts/braces to support, it would offset the supposed savings. In Calif where OP posts from, it is not at all unusual to see fences traversing the ruggedest of landscapes.

Four [4]

I don't know but I could ask my neighbor he is putting up a fence that looks like the old west forts.

C'mon Janissaries,

We know you've been reading the Thread you started.

How about advising us all and stating some of the things which at this time, are still variables.

Then proper assessment and calculations may be completed...

Talk about measuring over terrain, here's something slightly off topic that some might find interesting.

http://www.gdg.org/Gettysburg%20Magazine/measure.html

Hi Janissaries - It is about time you come back in defence of your fence.

I can recommend an uneducated farm worker for putting up the fence for you. He do not uses a tape or dumpy or anything but paces the posts. His fence always come out perfect.

The answer is 17 posts

As I said earlier, and other stated too:

The problem was really the "Woolly Specification", which allowed so many possible variables.

The original Question Specification was only stating a fence, posts, and a hill, with a paltry few numbers, assuming everything else.

Rather dangerous when there are enterprising Contractors around, who hope to get huge extras on the Contract, against a "Loose Specification".

Kind Regards....

Line of sight distance 1500 feet - requires 76 posts.

Terrain distance due to hill 1604 feet - requires 83 posts.

Ergo 7 extra posts are required.

The hill arc distance has a chord (L) of 225 feet - a median (M) of 100 feet

the radius of the 'hill circle' (r) = (L^2/8M)+(M/2) you can do the math to get the circumference and then the arc portion made up by the hill.

I've just discovered how to do the surveying bit, using a slide rule for the calculations (with apologies for the crap graphics):

Appendix - Tacheometric Surveying

STADIA COMPUTATIONS

In the instructions for Differential Trigonometrical scales, reference was made to the complete solution of the triangle given one side and one angle (other than the right angle).

Since the computations of heights and distances from stadia data call for the treatment of two overlapping right angle triangles, it will be seen from the following that a rule which incorporates the differential Sine and Tangent scales provides the means of effecting the evaluation of D, v and Z (see figure) when G and V° are known, by a single setting of the slide (occasional end switching excepted).

In the following it is assumed that the tacheometric instrument used is provided with an analatic lens, also that the measurement staff is held in a vertical position

Where

V° = the inclination of the telescope

S = the (vertical) staff reading = AC

r = the 'corrected' (staff) reading = A1C1

D = the horizontal distance

v = the vertical distance

Z = Distance on Collimation Line

G = Uncorrected distance Z = k S

k = Constant for the wires (usually 100)

e.g. Determine D, v and Z, when S = 4.5 ft and V° = 200 and the constant for the stadia lines = 100

G= 100 x 4.5= 450, V° = 20°, 2V° = 40°

Set the slide so that the 400 mark on the Sd scale aligns with 450 on the D scale

Then in alignment with:

20° on the Sd scale read the value of Z = 423 on the D scale

20° on the Td scale read the value of D = 397.4 on the D scale

20 on the C scale read the value of v= 144.6 on the D scale

That is, evaluate in accordance with the following diagram.

In the case of a value of G which occurs towards the left of the D scale such that Sd2V° when aligned with G results in Sdv°, Tdv° or Cv° being to

the left of D1, then end switching must be effected by placing the cursor at C10 and then bringing C1 to the cursor. This will enable the values of

the remaining projections on the D scale to be made.

The mathematical proof which substantiates the correctness of the foregoing method follows.

From the figure and remembering that

Sin V° Cos V° = Sin 2V° / 2

we get:

v = G Sin V° Cos V° = Tan V° G Cos2 V° = D Tan V° = Z Sin V° = G Sin 2V° / 2

Then

V°/ v = V°/ (G Sin V° Cos V°) = V° / (Tan V° G Cos2V°)

= V°/ D Tan V°

= V°/ Z Sin V° = 2V° / (G Sin2V°)

V°/ v = (V° / (Tan V°)) / D = (V° / (Sin V°)) / Z =(2V° / (Sin 2V°)) / G

i.e.

V°/ v = Tdv° / D = Sdv° / Z = Sd2v° /G = Sd2v° /(k s)

The slide rule interpretation of this form is given in the previous figure

When the angle V° is small (say less than 3°), then within the accuracy limits of measuring 'S'

(i) S » r

(ii) Z » l00r

(iii) D » Z

(iv) v » 100 r V°/ 57.3

e.g. Given V° = 1° 14' (i.e. less than 3°) and r = 6.17

To determine D and 'v'

Express 1° 14' as a degree and decimal = 1.233°

D » 100 x 6.17 = 617

To obtain 'v' evaluate

D V° / 57.3 = 617 x 1.233 / 57.3

Set cursor at D617, U (at C57.3) to cursor, cursor to 1.233

Read 'v' at cursor on D = 13.28 ft

You may need to visit http://www.remanley.clara.net/pdf/thornton.pdf for further clarification.

This is my slide rule (British Thornton model No: P271 - recommended by the school)

(For anyone else interested in slide rules, see http://cr4.globalspec.com/thread/2946/The-Million-Dollar-Slide-Rule - I missed it 1st time round).

Thank you John,

I haven't done stadia computations since 1963, and your Post has brought back memories.

Now I wonder where my old slide rule went, I still had it some 30 years ago.

Mind you, the question Specifications were rather "loose" as I mentioned earlier, because there were so many possible answers, all of which could have been correct.

Kind Regards....

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Look at the picture. Each verticle line represents a fence post. It takes the same number whether the ground is flat or going over a hill because the fence posts are installed vertically. The only thing that changes is the amount of wire used between posts.

It's cutting half the picture out but you can still see what I'm talking about with the half that is showing.