Geometry Question

Hi Again

This is posted in the General section. The answer depends on knowing the radius that generates the arc.

The radius , AX is the hypotenuse of a right triangle. AB is one leg of this triangle. Pythagorus said that AB² + BX² = AX² (Radius) Then BX = (AX² - AB²)^1/2

BD = RADIUS DBX - BX

TM

Hi

grab a copy of machinery's handbook and look up circular segments

Are you assuming that ADC is a segment of a true circle? And are you also assuming that AB and BC are not necessarily the radius of that circle?

Let arc length ADC = a, and let ABC = b, data for your problem.

Then define theta (in radians) and R so that a = 2*R*theta, and b = 2*R*sin(theta)

Then theta can be determined by solving sin(theta)/theta = b/a numerically (Use the "Solver" tool in Excel, or draw a graph, or try using calculus).

Then calculate R = a/2/theta.

Now the solution you require is DB = R*(1-cos(theta))

Regards,

Greg

(P.S. there is no 'tidy' solution because sin(theta)/theta is not one of the standard functions although it is tabulated numerically in some optics texts, where it is used in diffraction calculations)

Yeah, I go along with that. I make DB 1.65m, and theta 1.131 rad = 64.8°.

Cheers....Codey

I also agree, CB=1.65m, Theta=1.131rad, radius of circle=2.873m

George

Another thought overnight - (a/b)*sinθ = θ can be solved by iteration, on a spreadsheet or calculator. In this case a = 6.5, b = 5.2, a/b = 1.25.

Need to work in radians of course. Put θ = 1, take sin, x 1.25 =, take sin again and so on. Gives θ = 1.13 after about 8 steps, 1.1311 after 17.

When θ is known, the other lengths can be found easily.

Codey

Codey,GREAT answer. I am not a math major or engineer but I love working with it. Your solution was simple and brilliant. Well at least I could understand how and why it worked.

MusicAl

Thanks Al, glad you liked it......Codey

db=(approx.)=(ab*ab)/R

R = diameter/2

AB*BC=DB* BE

AB*BC=DB* DE-DB_a

2.6*2.6

BE= DE-DB

I assume DE is diameter in these formulas (property of intersecting chords) but we don't know diameter up front.

It's not clearly stated that the drawing represents part of a circle. I haven't worked it through , but it looks possible from the value you calculated. I won't confuse things but if you want to glimpse what I mean , look here. The solution for a cycloid would be easy.

Yes, you're right. It looks like it on sketch but mustn't make assumptions. Clearly type of curve must be known to calculate DB.

I'm not too sure about your comment "The solution for a cycloid would be easy". Can a cycloid be specified to give these lengths? To get the right AC, dia of the generating circle = AC/pi = 1.655, but how do we know this gives the required arc length? I first worked it out the other way round, and got dia = 1.625 (exactly) to give arc length 6.5. But 1.625*pi = 5.1, not 5.2 for AC.

Maybe part of an ellipse could be tailored to suit, I suspect from a quick think there might be an infinite number of them.

Cheers....Codey

Hi codey , I'm in a bit of a rush right now. If it were a cycloid the arc length would simply be 2* pi*r where r= (oops can't see notation in editor)

Draw the path of a point on a circle as you roll it along a line , and you will see known dimensions. I will check your previous answer , it looked (initially to be the right size). Back later today hopefully. Kris

Codey. The curve is not cycloid. If it were though , you would have

BD = 2*r (generating circle )

ACD =2*Pi*r

ABD=8r

The Known ABD and ACD are not consistent with this.

Your conclusion of #11 was sharper than mine.

It's an interesting factor to consider though (what other curves fit ). Kris

Kris - exactly, that's the point I was trying to make.

Cheers....Codey

Hi Kris, I've had another look at this (sad life, bank holiday weekend!) and I'm not so sure.

Your discussion, and mine in #11, assume the base line of the cycloid is line ABC. But it doesn't have to be, (and if it were, it would hit points A and C at 90°). Possibly there is a cycloid, generating circle radius > ABC/2pi, which satisfies the given conditions. I don't think I'll bother trying to work it out though.

Also I'm not sure whether, if we're told it's part of a cycloid, this fixes both the generating circle radius and BD. If not, the question would also need to give the generating circle radius.

Cheers....Codey

"Archimedes! What is zit with you! Always with za triangles! Not everything is about triangles!"

- Albert Einstein to Archimedes

- Red Dwarf

When I slot you into a console game you're gonna get real triangulated. You may end up with Lisa Simpson hair if I'm feeling lazy.

When you have been given two values - the length of arc ADC, and the length of the segment AC, I believe you need to specify the radius of the circle to calculate DB. Without that you can get different values of DB for different values of the radius. (I guess this point might have been brought out in the numerous expressions in the replies you have received.)

Hello All,

Let me clarify this in simple terms.

The Arc ADC will always be variable in lnegth .

The Line ABC will always be variable.

What I need to know is a close approx. of what the distance is between DEAD CENTER of the arc and DEAD CENTER of the Line.

Thank you all for the formulas and such. I am not an engineer so please keep in mind my math skills are limited in this area.

netmaker

But the type of curve is important (see various postings) and you haven't mentioned this. Can you confirm it's arc of a circle? If not, what is it? If it is, my answer is in post #6, for the lengths in question (trust others agree?).

For other lengths, Guest #4 gives a more general approach, but as he says it needs a trial solution of some kind, as in general equation with both angle and sin(angle) can't be solved directly.

Cheers....Codey

It can be any thing between 1/2*ABC and 0

You need to know the radius

Can't agree Dovrak. Assuming it's arc of a circle, dimensions given fix the circle radius.

If BD = 1/2*ABC = 2.6, it's a semicircle, radius BD. This gives arc ACD = pi*BD = 8.17, but figure in original post = 6.5.

If it's 0, BD = AB and radius is infinite. ADC = ABC = 5.2 (vs 6.5).

Actual radius is somewhere between, I make it 2.873 (BD = 1.65 - see #6)

Codey

Draw a circle with chord ABC.With values given,and radius R and theta subtending arc ADC get the following relationships: 6.5=R theta; cos theta/2=2.6R. Eliminating R we get cos theta/2=.4 theta.UnfortunatelyI do not have the computer tools to make the relationship converge.I solved for theta by trial and error.I get theta=95.93 deg with .02%error.Solving for R I get R=3.883m By trig I get center to B=2.6m subtracting this from 3.88m I get BD=1.28m. Now knowing R and theta check S=R theta I get R=6.501m.

Unfortunately, I do not see a closed form equation.

netmaker's question is in regard to what is/was generally known as the Sagittarius or archer's bow problem, finding the altitude of a segment of a circle.

Have you considered using calc to solve this variable arc problem.

The Circumference is = 2piR where R = BD + r. r=the variable incremental distance to the center of the arc at apoint we call F for foci. The distance DF =R. The distance DB= DF -r.

Since we know the area of circle is the integral of the circumference we might be able to get close enough using some calc 101 stuff.

The Area of the circle is piR^2, that is pi R squared. It also happens to be the integral of the Circumference.

In general terms if you integrate 2piR it equal piR^2.

What you also know is the R is 1/2 the distance of ABC plus some extra distance r. Therefore R = 2.6 +r.

Substitute in the above formula and sove for r using quadratic equation.

Area=pi(2.6+r)^2 = pi6.76 +5.2r +r^2

Circum = 2pi(2.6 + r) = 5.2pi + 2pir.

Integrate Circum and get 5.2pi r + pir^2

Set them equal to each other 5.2pir + pir^2 = 6.75pi +5.2pir +pir^2

Move all the like terms together set equal to zero and solve for little r using quadratic equation.

coefficients a=1, b=3.2 and c=1.56?

check math here for me.

Then solve for R by adding 2.6 to the little r.

So you have a piece of material 6.5 m long and you'll fix the 2 ends at 5.2m to form an arc. You want to know how high it'll hang?

Use the calculation posted, assuming its a circular arc. Actual height will be less due to weight of material. The heavier the material, the lower it'll be.

Pineapple

There is only 1 solution to the problem if this is the arc of a circle. I feel that this chord is not a diameter but rather a Chord of distance 5.2 meters and Arc length of 6.5 meters. If chord is not a true diameter then radius is not 1/2 ABC. If you use a tape to measure the arc length, then point D can be marked and a true measurement taken. If all is theory then I suggest that you get a surveyors field book. all of your trig formulae and curve formulae are inside the covers of the book. These run about $7 to $10 depending on size desired.

the solution to this problem is as follows:

Delta = 129°36'53"

R= 2.87 meters

Arc length = 6.50 meters

Chord length = 5.20 meters

Tangent Length = 6.11 meters

External = 3.88 meters

Middle Ordinate = 1.65 meters (Your requested value)

Any good survey program can calculate these values if two parameters are known.

P C Survey has a COGO (coordinate geometry) screen for about 200 dollars that will allow you to calculate these values for any curve if you know two (2) parameters. I Chord length and Arc length and did this in about 30 seconds with my COGO screen.

Good Luck

BRAVO!!!!

trasfer the geometry problem on the cartesian system which would be easiest to study it ..AC is a space on the xx' and Ab=Ac/2 and take that the arc AC(name it F) is a equality of AC=> F(AC)=.. take the F'(AC) and then the F''(AC) the spot were F''(AC) changes from possitive to negative, is the spot D(x,y)(actually is the local max of F(AC) the x is always B's x and y is the height of D.

Nice work. but what values do you use to establish AR when solving for BR? You still need to find the radius.

I think there are TWO unknowns, so we need Two equations. Like Circum and Area. Both have radius in them. What did you think of the calc approach?

Dear netmaker ,

It is easly to find that :

DB = R - [R² - (ABC/2)²]^0.5 ...................................................... (1)

Sin Θ = BC / R ........................................................................(2)

where R = Radius & 2Θ = Arc Angle in degrees

2Θ / 360^o = ADC/ 2∏ R , where ∏ = 22/7 (or 3.14)

Θ = 90^o (ADC / ∏ R) ..............................................................(3)

By dividing eqs. (2) & (3) : Sin Θ= ∏ (Θ/90^o)(BC/ADC) ..............(4)

In curve drawn by eq. (4) by substituting for Θ and calculating sin Θ, you will find that there is only one value on the curve for sin Θ which is equal to the true value.

For BC = 2.6 M & ADC = 6.5 M :

Θ = 64.80739963^o

R = BC/Sin Θ = (5.2/2) / Sin 64.80739963^o =2.873302711 M (from eq. 2)

DB =R-[R²-(ABC/2)²]^0.5=2.873302711-[2.873302711²-(5.2/2)²]^0.5= 1.650245693 M (eq. 1)

The only value for Θ on curve drawn by eq. (4) is -ofcourse- that value of Θ ≠ 0.

theta=l/r ,sin(theta)=x/r,sin(theta)/theta=x/l; where r=radius,x=AB;l=AD;

now,knowing the value of theta from above two equations

use r^2=(r-y)^2 + x^2 ;where y=BD..

from these calculate bd...