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Topic: My D.I.Y. Cobra Drift Anchor  (Read 73303 times)

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polepole

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The volume of a tetrahedron  (pyramid with a equilateral triangles on all sides) is equal to sqrt(2) * (a^3) / 12, where a is the length of an edge.

The volume of a pyramid with a square base and equilateral triangles as sides is equal to sqrt(2) * (b^3) / 6, where b is the length of an edge.

Solving for the relationship of a to b assuming equal volume of the two pyramids yields.

sqrt(2) * (a^3) / 12 = sqrt(2) * (b^3)  / 6
              (a^3) / 12 =               (b^3) / 6
              (a^3)        =               (b^3) * 2
               a             =              ((b^3) * 2)^(1/3)
               a             =                b       * 2^(1/3)
               a             =                b       * 1.2599

Therefore the leading edge of a tetrahedron is 1.2599 times the leading edge of the pyramid with a square base and equilateral triangles for sides.

However, this assumes the triangles are equilateral.  My head is hurting from this one, is yours?  What if the goal was to make sure the angles of the sides of the pyramid were the same for both cases, so the pyramids would sit at the same angle on the bottom.  Would the sides still be equilateral for pyramids of the same volume?  I'll leave this to the "next nerd" to do the analysis.   ;D

-Allen



Spot

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FYI-my yak anchor only weighs 9 pounds and it has been fine.

Man, I couldn't hold a good spot with 10lbs.  If I scoped too much rope, I wound up on a merry-go-round down current and then back up the eddy.  I guess I'll have to get in some more practice soon.  Bummer  ;D
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.  --Mark Twain

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Tournament Results:
2008 AOTY 1st   2008 ORC 1st  2009 AOTY 1st  2009 NA Sturgeon Derby 1st  2012 Salmon Slayride 3rd  2013 ORC 3rd  2013 NA Sturgeon Derby 2nd  2016 NA Chinook Showdown 3rd  2020 BCS 2nd   2022 BCS 1st


steelheadr

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The volume of a tetrahedron  (pyramid with a equilateral triangles on all sides) is equal to sqrt(2) * (a^3) / 12, where a is the length of an edge.

The volume of a pyramid with a square base and equilateral triangles as sides is equal to sqrt(2) * (b^3) / 6, where b is the length of an edge.

Solving for the relationship of a to b assuming equal volume of the two pyramids yields.

sqrt(2) * (a^3) / 12 = sqrt(2) * (b^3)  / 6
              (a^3) / 12 =               (b^3) / 6
              (a^3)        =               (b^3) * 2
               a             =              ((b^3) * 2)^(1/3)
               a             =                b       * 2^(1/3)
               a             =                b       * 1.2599

Therefore the leading edge of a tetrahedron is 1.2599 times the leading edge of the pyramid with a square base and equilateral triangles for sides.

However, this assumes the triangles are equilateral.  My head is hurting from this one, is yours?  What if the goal was to make sure the angles of the sides of the pyramid were the same for both cases, so the pyramids would sit at the same angle on the bottom.  Would the sides still be equilateral for pyramids of the same volume?  I'll leave this to the "next nerd" to do the analysis.   ;D

-Allen


I need some help with all that math.

Oh, here it is...  :ky:

 :D
"Fast enough to get there...but slow enough to see. Not known for predictability"  Thanks to Jimmy Buffet for describing my life...again



SBD

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I have used the same anchor on my X-Factor, Mini-X, and my T11. It ought to hold.


ZeeHawk

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FYI-my yak anchor only weighs 9 pounds and it has been fine.

Man, I couldn't hold a good spot with 10lbs.  If I scoped too much rope, I wound up on a merry-go-round down current and then back up the eddy.  I guess I'll have to get in some more practice soon.  Bummer  ;D

I'm guessing you anchored w/ the Fish and Dive? Since SBD is fine w/ only 9# could be possible that anchored from the stern the F&D isn't as hydrodynamic as his X-Factor, Mini-X, and T11? Therefore the drag is pulling you downriver? Just a thought.

Z
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Pelagic

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20lb holds my fnd.. I don't think 9 would cut it, at least not in moderate speed moving water. River bottom composition has a lot to do with it too.  I also know the fnd is not the most hydrodynamic of boats when anchored from the stern. It's lack of rocker in the stern make me a little nervous and careful where I drop the anchor.  I always use plenty of scope and stay away from heavy current. I just use it to hop from hole to hole then either sit sideways and fish or get out and wade.


Spot

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FYI-my yak anchor only weighs 9 pounds and it has been fine.

Man, I couldn't hold a good spot with 10lbs.  If I scoped too much rope, I wound up on a merry-go-round down current and then back up the eddy.  I guess I'll have to get in some more practice soon.  Bummer  ;D

I'm guessing you anchored w/ the Fish and Dive? Since SBD is fine w/ only 9# could be possible that anchored from the stern the F&D isn't as hydrodynamic as his X-Factor, Mini-X, and T11? Therefore the drag is pulling you downriver? Just a thought.

Z

You could be right.  There's not a lot of rocker in the tail of my F-n-D and I can really feel the effects of the keel when trying to change lines in the fast water.  It might also have something to do with the volume of water that moves thru the Nestucca.  Even the soft water is moving pretty fast down that river.  

I'll have to get together with INSAYN to fab a Trident drift anchor and maybe pour a 15lb pyramid for the F-n-D.
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.  --Mark Twain

Sponsors and Supporters:
Team Daiwa        Next Adventure       Kokatat Immersion Gear

Tournament Results:
2008 AOTY 1st   2008 ORC 1st  2009 AOTY 1st  2009 NA Sturgeon Derby 1st  2012 Salmon Slayride 3rd  2013 ORC 3rd  2013 NA Sturgeon Derby 2nd  2016 NA Chinook Showdown 3rd  2020 BCS 2nd   2022 BCS 1st


INSAYN

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Allen - Thanks for the math class, now my head hurtz.  ;D

Ya'll, what would be the longest measurement a person would want their anchor to be from hanger loop to pointy tip, while dangling in the air? 
 

"If I was ever stranded on a beach with only hand lotion...You're the guy I'd want with me!"   Polyangler, 2/27/15


PNW

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The volume of a tetrahedron  (pyramid with a equilateral triangles on all sides) is equal to sqrt(2) * (a^3) / 12, where a is the length of an edge.

The volume of a pyramid with a square base and equilateral triangles as sides is equal to sqrt(2) * (b^3) / 6, where b is the length of an edge.

Solving for the relationship of a to b assuming equal volume of the two pyramids yields.

sqrt(2) * (a^3) / 12 = sqrt(2) * (b^3)  / 6
              (a^3) / 12 =               (b^3) / 6
              (a^3)        =               (b^3) * 2
               a             =              ((b^3) * 2)^(1/3)
               a             =                b       * 2^(1/3)
               a             =                b       * 1.2599

Therefore the leading edge of a tetrahedron is 1.2599 times the leading edge of the pyramid with a square base and equilateral triangles for sides.

However, this assumes the triangles are equilateral.  My head is hurting from this one, is yours?  What if the goal was to make sure the angles of the sides of the pyramid were the same for both cases, so the pyramids would sit at the same angle on the bottom.  Would the sides still be equilateral for pyramids of the same volume?  I'll leave this to the "next nerd" to do the analysis.   ;D

-Allen
thought that might be the case, but started to smell burning insulation when trying to figure out the math. thanks Allen

hey now... did you get a chance to figure out why recent pics aren't showing up in the gallery?  ??? :angel12: ;D tried uploading again today from Foxfire & IE... no luck. I'm wondering if somthins wrong w/ my computer. anyone else having this problem?


Spot

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Ya'll, what would be the longest measurement a person would want their anchor to be from hanger loop to pointy tip, while dangling in the air? 

I'll venture a guess here and say that the critical limitation (beyond the fact that increased height would result in decreased length of "holding edge") would be the distance from the pulley to the surface of the water.  Only bad things can come from your anchor dragging in the water.
It appears that all the pyramid anchors I've looked at are equilateral triangles   
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.  --Mark Twain

Sponsors and Supporters:
Team Daiwa        Next Adventure       Kokatat Immersion Gear

Tournament Results:
2008 AOTY 1st   2008 ORC 1st  2009 AOTY 1st  2009 NA Sturgeon Derby 1st  2012 Salmon Slayride 3rd  2013 ORC 3rd  2013 NA Sturgeon Derby 2nd  2016 NA Chinook Showdown 3rd  2020 BCS 2nd   2022 BCS 1st


polepole

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Am I going to have to toss back a couple shots of scotch and do the analysis or is some other math geek going to chime in?

-Allen


bsteves

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Why are we doing all this math again?  Are we trying to reinvent the pyramid anchor?

As for the couple of shots of scotch... mmm scotch.  I think I'll get myself a dram of my Ardbeg.
“People say nothing is impossible, but I do nothing every day.”

― A.A. Milne, Winnie-the-Pooh


INSAYN

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Why are we doing all this math again?  Are we trying to reinvent the pyramid anchor?


Because we can!  :icon_thumright:

Ponder this:
If we didn't expect folks to "reinvent" a proven design, we would all be paddling logs instead of the variety of kayaks we have now.  We'd still be fishing with a bamboo stick, dental floss, a rock, and a safety pin. :o
 

"If I was ever stranded on a beach with only hand lotion...You're the guy I'd want with me!"   Polyangler, 2/27/15


 

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