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Drawing circle with diameter 7917 miles to find drop

trickyK3C2L
Explorer

Drawing circle with diameter 7917 miles to find drop

trickyK3C2L
Explorer
Explorer

Hey guys & girls I'm not an Autocad guy and I've downloaded a trial version to see if I can do the drawing and figure out the math BUT its harder than I thought. 

I wanted to draw a circle with a diameter of 7917 miles, then at the top draw a 2 mile long line that starts and finishes on the circle, then measure the drop of the line at points 1/4 mile, 1/2 mile, 3/4 mile to see how far the drop is in inches.

is this possible to be done in Autocad.

Thanks in advance 

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37 Respuestas
Respuestas (37)

Kent1Cooper
Consultant
Consultant

Draw a LINE from the top-most QUAdrant point of the Circle, downward by whatever distance [it won't need to be very long].  OFFSET it by one mile to each side.  Draw your 2-mile chord Line between the intersections of the two Offset results with the big Circle.  Offset or COPY the original Line by 1/4 mile distances to get your drop locations, and TRIM the results so they run only between the horizontal Line and the Circle.  The lengths of those will be the drops at those locations.

 

[The size of the big Circle looks like it's a circumference of the Earth.  If that's what it is, I assume you are aware that Earth is not perfectly spherical, but bulges outward some at the equator from the centrifugal force of rotation.  So your drop distances will be smaller near the poles than near the equator.  And of course, there is probably no place on earth where the surface perfectly conforms to that curvature for a distance of 2 miles.  I confess to being curious about the purpose....]

Kent Cooper, AIA

trickyK3C2L
Explorer
Explorer

Hey ken thanks for the reply, I will try it but to be honest I dont think I will be able to do what you mentioned as its too much for a newby to understand, but I will give it a go.

LOL yes it is the size of earth and Im wanting to put this argument to bed finally with mechanical test.

After 2 years of laser and camera tests I cannot find any curvature that exists on this spinning space marble, so Im on my way to do a mechanical test where i will span a top quality fishing line across a 2 mile dam with approx 200 lbs tension just 1in above the waters surface at both ends, and if the earth is a globe it will drop into the water by some distance (reason for autocad). The normal curvature at 2 miles is 32in, but I want to know what Autocad will show through the middle when I have the fishing line 1in above the surface, as it will be less than 32in.

Its been done just recently, but I want to do it here in Australia also to prove as Fact.
https://youtu.be/xQ7fxn5dym8?si=ciuNwyVH7wi6FdgS


If you can help me draw it that would be great, as it would be good to know what the drop would be at what distance and where its the greatest drop.

Email me if its easier

Thanks heaps 


Tricky

 

@trickyK3C2L - this post is being edited to remove PII.

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Washingtonn
Collaborator
Collaborator

If you spent 2 years testing with lasers and cameras without finding any curvature to the earth, you should have no qualms about using the method of Eratosthenes used over 2000 years ago to proving curvature of the earth using a couple of sticks and some guys pacing out the distance between 2 cities. The advantage of modern GPS and other precise measuring devices, should allow you to develop a test setup without the hassles of using a fishing line.

 

The experiment you linked did not account for the catenary of the fishing line - a major factor to be considered in generating an accurate measurement.

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trickyK3C2L
Explorer
Explorer

Hey matey thanks for the reply, I appreciate your time, just need help as I still cannot navigate this program lol.

In relation to the test you mentioned (Eratosthenes) to do this test, I have to assume a distant sun with parallel sun rays, and this is not what we experience here where we live, and assuming a parameter in a test without proof will just void the test immediately.
Both models work with his test as the flat earth has a local close sun, giving the same answer with shadows due to the sun location, and we experience corpuscular sun rays here with no sign of parallel sun rays, so defiantly moves towards flat earth.

Anyway I know the test I do will finish any arguments as if we live on a globe the fishing line "MUST" drop under the water surface due to earth curve, so even with your (catenary of the fishing line) this will drop the line into the water and help the globe model, so if I find the fishing line above the water "At all points over the 2 mile", even with the line drop (not much at 200lbs tension), then the answer is the water is level & flat & horizontal, and the earth is also flat.

Can you help me please with maybe drawing it for me and sending a file over I can look at, as man o man you guys are good as I have no idea how to do it lol.
Thanks heaps 

Tricky

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Washingtonn
Collaborator
Collaborator

The sun rays are essentially parallel everywhere the sun shines on the earth.

Check your calculations for what the drop over 2 miles is if the earth is a globe.

Check out this link which showing that the test Eratosthenes method does in fact work down under:

https://www.researchgate.net/publication/272755423_Modern_replication_of_Eratosthenes'_measurement_o...

Save yourself some money and time  - abandon the fishing line test just go to the coast and watch as ships sailing away disappear below the horizon line -they aren't sinking nor falling off the edge of the earth.

 

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trickyK3C2L
Explorer
Explorer
lol,

This is the issue, people don’t realise what they are seeing,
I’ve been as far as 14 miles at 1 ft above the water and I can see houses on the opposite side that should be well over a curve,

Drop at 2 miles is 32in, with a straight line horizontal, but I need help to work out the amount a straight line will be below the water each 1/4 mile,
Can you help me please
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richard_387
Advocate
Advocate

I can understand your reticence to use AutoCAD with these very large dimensions. It is simple to draw once you have confidence in what is happening, and there is a problem presenting the results.

 

I drew it using a drawing unit to be an inch. So the very first thing is to convert the diameter into inches - 501,621,120 inches. Then there is a problem presenting the results because of the scale  of dimensions involved. I have presented a pictorial image which shows what information you are after.

 

fishing.PNG

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leeminardi
Mentor
Mentor

I would start by detemining the theoretical mid-length sag of a cable.  The standard engineering mechanics equation for the sag of a catenary curve (i.e., a freely sagging cable) is:

 

sag = (T/W) *[cosh( (W*L)/(2*T) )- 1]

 

where

T= mid-span cable tension

W = weight per unit length of the cable

L = distance between the supports

 

A 250 lb monofilament fishing line has a diameter of 1.6mm. Assuming a density for nylon of 0.036 lbs/in^3 yields a weight/inch of 0.003116 lbs/in for the fishing line.

 

If the attachment points are 2 miles apart the midspan sag is about a half-mile!  Here are the more precise calculations via Excel.

leeminardi_0-1728917515337.png

How should the sag be measured? From a straight line between the two end points or relative to a curve with a radius of Earth?  In the real world example it would be relative to the arc of the Earth. 

 

If you'd like to know more about the shape of the earth I recommend:

Measure of the Earth: The Enlightenment Expedition That Reshaped Our World

 

It's a surprisingly interesting story.

 

 

 

lee.minardi
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richard_387
Advocate
Advocate

What happens when each end of the fishing line were to be placed at water level or slightly under? The density of water appears to be similar to nylon, so the line would not have the same downward force as being in free air.

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dbroad
Mentor
Mentor

Draw a 2 mile radius circle centered at the top quadrant of the earth circle.  Then draw a line from the quadrant of the earth to the intersection of the two circles. Divide that line into 8 parts with the divide command.  At each point, draw a line from node snap to per of circle.  Select or dimension each line to see the length.

 

Note though the earth really isn't circular so the results will not be precise to the nth degree.

Architect, Registered NC, VA, SC, & GA.
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trickyK3C2L
Explorer
Explorer

Hey Richard thank you so much for the help, as you can imagine I was stumped on how to draw it & see the results lol.

I knew it was 8in per mile squared off a horizontal line down to earths surface, but I wanted to know in CAD how it would be when used on a assumed globe, and to be honest I thought it would be closer to the 32in drop.

 

I will use this drawing for my test, and I will use braided line as it has no stretch & use around 250lb stretch to make sure when its pulled to tension @ 1in above the water surface, it will show the results to finally put this argument to bed, and stop all the divide between everyone.

Thank yo for your help its much appreciated,

 

Send me an email & I will keep you up to date when I carry out the test.

 

Thanks again

 

@trickyK3C2L -  this post is being edited to remove PII.

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trickyK3C2L
Explorer
Explorer

Hey Lee I will be using braided line for the test as its smaller & HAS NO STRETCH.

 

I will do the first test 1in above the waters surface, and if the line stays above the water all the way across the dam then the earth is flat, and again if the line drops under the waters surface and becomes buoyant, I will raise the line to 8in above each end, and see the drop in the middle.

 

From my pictures & laser tests carried out already I know the result, just want to prove it to stop the argument.

 

Thanks again

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trickyK3C2L
Explorer
Explorer

Thanks again guys for your help, I will repost once experiment is carried out.

Just trying to gain access to the dam for the test.

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trickyK3C2L
Explorer
Explorer
Hey Richard thanks again,

I was thinking if we drew a line starting at the water surface (at mid point), and it never went under the water, does Autocad show approx 32in above water surface at the 2 mile distance.

Thanks
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leeminardi
Mentor
Mentor

@trickyK3C2L wrote:

Hey Lee I will be using braided line for the test as its smaller & HAS NO STRETCH.

 


Braided Nylon WILL stretch.

leeminardi_0-1728949781061.png

A quick search indicates that braided nylon has a modulus of elasticity (a.k.a. Young's modulus) of about 0.8 x 10^6.  You can cacluate the stretch with:

 

   Change in length = ( force * orginal_cable_length) / ( cross_section_area * modulus)

 

To get a sag of only 1" I think you will need a force much greater than 200 pouds. Perhaps over 10000 pounds for a mile long wire. It depends on the cross sectional area and the length.

lee.minardi
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trickyK3C2L
Explorer
Explorer

Thanks matey,

 

I will run the test and video my results to see what I get & if I acquire any sag of noticeable amount, and i will know real quick once load is placed on the line, as ill start at 50 lbs & move up at 50 lb increments, 
I just found out they offer a 500 lb line for the massive game fishing, so await specs to see if better to use.

 

thanks again  

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richard_387
Advocate
Advocate

I have redrawn the picture to show the dips when starting at the surface at one end.

 

It does indeed show a 32 inch distance at 2 miles.

 

fishing2.PNG

trickyK3C2L
Explorer
Explorer

Thank you Richard,

proves the 8in per mile squared works perfectly,

I heard its accurate up to 1000 miles, then the parabular formula is out.

Appreciate it,

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Kent1Cooper
Consultant
Consultant

@trickyK3C2L wrote:

.... Can you help me please with maybe drawing it for me and sending a file ....


See the attached, without the vertical exaggeration of @richard_387 's images, and ignoring the fact that the Earth is not actually spherical.  A drawing unit is an inch.

 

The red is my suggested [in Message 2] initial Line from the top quadrant point downward, the yellow are the 1-mile offsets to each side, the green is the chord over that 2-mile distance, and the white are the quarter-mile-interval drops [you'll need to Zoom way in, and maybe REGEN].

 

You talk of a string with no stretch, but even if that could possibly be true, that doesn't mean it has no sag over a spanned distance.  To actually get zero sag would require infinite tension, no matter what the material.  And of course, at this degree of curvature over this kind of distance, any little ripples / waves in the water will spoil your test.

Kent Cooper, AIA
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