My D.I.Y. Cobra Drift Anchor

[polepole]

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.   

-Allen

[Spot]

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  

[steelheadr]

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.   

-Allen

I need some help with all that math.

Oh, here it is... 

 

[SBD]

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

[Zee]

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  

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

[pelagic paddler]

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]

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  

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.

[INSAYN]

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

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? 

[PNW]

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.   

-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?  tried uploading again today from Foxfire & IE... no luck. I'm wondering if somthins wrong w/ my computer. anyone else having this problem?

[Spot]

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   

[polepole]

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]

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.

[INSAYN]

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

Because we can! 

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.