Lanyard Bead Dimensions (Help Me Out!)

Okay, today I need YOUR help…

Due to popular demand in a previous post I did – I’m going to produce some titanium lanyard beads (the titanium is on it’s way to me as I type this!).

I’ve got the designs already for the most part (I’ve had them for well over a year – but they’ve been low-priority).

Here’s where you come in

I need to get a better idea of dimensions before final design and creation.

Now, I could go and do  lot of research – but I’m confident enough that I can trust what you tell me.

Because, almost certainly, you are way more knowledgeable about lanyard beads than I am.

There are two dimensions I need to know – take a look at the drawing below:

Lanyard Bead Dimensions

Because the designs are 90% done – I just need to finalise ‘L’ (the overall length of the beads) and ‘H’ (the internal hole that the cord passes through)…

What should ‘L‘ be?

What should ‘H‘ be?

Now, I expect some differences of opinion.

That’s fine.

Mainly looking for a consensus of what’s going to work for most.

Thank you in advance!


Oh, and one more thing

The beads are going to be the usual Cogent Industries high-end design and quality – so they’re not going to be low-end $5-each type things.

The machining costs are relatively high for such small objects – but the designs are pretty damn cool.

You know me… it’s either awesome …or nothing!

I vote for awesome – how about you? 🙂

  • Jon says:

    There’s a few out there. Most of mine are by Pete Gray. ID is .200 +/- .050 across my collection. Starlingear beads are weird, not always circular hole.

    Length should be at least the OD of the piece. Maximum 2x OD IMHO.

    Waiting for the chance to get one. Anodized them.

  • Tammy says:

    Hole should definitely be 6mm. Lenght should be about 15mm. But depends on the design of course.:D

  • Peter says:

    H=(2 x diameter of paracord 550) – thus allowing it to function as pacing beads.
    If that ole is too large then H should = (2 x diameter of paracord 550 with inner strands removed)

    L = 12 or 15 mm..

  • Dwayne says:

    Hole= 6 Millimeters
    Length = 13 Millimeters


    Hole= 1/4.Inch
    Length= 1/2 To 5/8 inch

  • William says:

    I think it should be L15mm and ID about 4-5mm, enough to be able to have two medium size cords through it without it being tight??

  • Gusto says:

    Those sizes seem right, but i would prefer a tighter fit versus a looser fit for the ID or H measurement. Overall length of .5 to .625 inches sounds good too.

  • Sazad says:

    Hole= H= 4.7mm ideal for 550 paracord x2
    Length = L= 22+mm I prefer longer beads easier to pull and hold

  • Darryl says:

    I believe that a 5mm internal hole dimension hits the sweet spot, not loose but not too tight.

    The length can vary, but I like it around 10-12mm depending on design.

  • David Smith says:

    Should make them similar to the ones on the viper fish. I find those a perfect size and fit. Don’t fix what isn’t broke 🙂

  • steve says:

    Here’s a site found that list diameter of various para cordage.

    Not sure if type ii or iii 550 is most common. 2x the most common for diameter. For length, 4x?


  • steve says:

    Here’s another site. Looks like 550, type iii is the one to design for:

    6mm by 18mm ?

    Need prototypes to play with 🙂

  • I would say the L somewhere between 15-20 mm
    The hole should be big enough to put a double para cord through… so 5-7/8 mm??

  • Jerry says:

    I have two sizes : 12.7mm (L) / 5.05mm (H), and 17.25mm(L)/5.85mm(H).
    Paracord with no inner strands fits the large one but not the small one.
    I prefer the smaller one and just use smaller cord.

  • Tom Anderson says:

    H = 4.5mm (minimum) – 5mm (maximum) & L = 1 – 2.5 X H

  • Man, fantastically detailed replies.

    Thank you so much to each of you (and to those who replied via email)

    From this I’ve pretty much got the dimensions I need – awesome.

    The titanium should be here any day now …and I’ll be looking to get it machined right away (the designs are pretty much done)

    Thank you again for the tremendous replies!

  • >