Merlin-like IG pool cover brass anchors on Amazon

Bugman1401

Well-known member
May 6, 2022
75
Charlotte, NC
I had an existing Merlin IG pool cover with the Merlin brass anchors mounted in the concrete pool deck which I’ve used for years. Last year I had the concrete pool deck demolished and replaced with travertine that looks great. Since pool season is now over I would like to cover my pool, but the original brass anchors went with the old pool deck. I’ve looked online and the Merlin brass anchors seem to be priced at $14-$18 each. There seems to be copy versions at $45 for a pack of 20 on Amazon.
Anybody have any experience with these copied versions and know if the quality is as good?
 
How thick is the travertine and how was it installed? Are the travertine tiles mortared to a concrete slab? You may need longer anchors since travertine is softer than concrete.
 
My travertine pavers are 1.375” thick and are not cemented down except for the coping which is cemented to the framing for the vinyl pool. However, the spring straps from my existing Merlin pool cover go way past the coping to the adjacent travertine that is locked in with all the other travertine pavers on compacted dirt.

Was looking at the brass anchors below. Suppose I’ll need a core drill bit.

 
A core drill will minimize the chance of cracking and chipping the tiles. Ideally it would be nice if the anchor could be located at the joint between tiles. Do you know how deep your pool lines are? Or are they run where they won't accidentally be hit. I've been using a core bit to go through the pavers then switching to a hammer bit to go through the slab under the pavers. To make starting the core bit easy, drill the same size hole in a piece of wood (2 x4 or a smallish piece of plywood). Stand on the wood and use it as a guide to get the core bit started.
 
My travertine pavers are 1.375” thick and are not cemented down except for the coping which is cemented to the framing for the vinyl pool. However, the spring straps from my existing Merlin pool cover go way past the coping to the adjacent travertine that is locked in with all the other travertine pavers on compacted dirt.

Was looking at the brass anchors below. Suppose I’ll need a core drill bit.


You don't need a core drill for these. Just a hammer drill with a masonry bit. A new bit will cut like butter. Get a couple bits.
 
You don't need a core drill for these. Just a hammer drill with a masonry bit. A new bit will cut like butter. Get a couple bits.
I would not hammerdrill travertine.

If the stones are not mortared down, then you should use paver tubes.

The anchor gets pulled by the spring from the back, which causes a rotation around the green pivot point, which causes the stone on the right to be pushed backwards and lifted up.

A cover cannot be safety rated without the paver tubes when set in pavers that are not mortared down.

1695561467290.png
 
Last edited:
  • Like
Reactions: Newdude
I would not hammerdrill travertine.

If the stones are not mortared down, then you should use paver tubes.

The anchor gets pulled by the spring from the back, which causes a rotation around the green pivot point, which causes the stone on the right to be pushed backwards and lifted up.

A cover cannot be safety rated without the paver tubes when set in pavers that are not mortared down.

View attachment 532145

Oh this is through travertine. Yes agree you need the paver tubes. I have a couple where the anchor is in the grass area. They go into the ground I think 24".
 
  • Like
Reactions: JamesW
Even if you go into the center of a paver, there is still a lifting force due to the application of a force above the surface of the stone.

The pivot point just gets moved to the edge of the stone.


1695562317341.png
 
1695566634447.png

If the strap tension is 40 pounds (No cover loading) and the anchor is 18” from the edge and the spring is 0.75” above the surface.

F = 40 pounds.

d = 18”

h = 0.75”

x = hypotenuse

X^2 = 18^2 + 0.75^2

X = sqrt(18^2 + 0.75^2)

X = 18.0156182242

F(lift) = 40(0.75)/18.0156182242.

F(lift) = 1.665 lb

F (horizontal) = 40(18)/18.0156182242

F (horizontal) = 39.965

Interior angles are (90° | 87.61° | 2.386°)≈(1.571 rad | 1.529 rad | 0.04164 rad)

We have a 40 pound force being applied at 2.386°.

The vertical component is (Sin 2.386°)40 = 1.665 lb.

The horizontal component = (Cos 2.386°)40 = 39.965.
________________________________________________________________________________________________________________________

If the strap tension is 100 pounds (Cover loaded with snow, water etc.) and the anchor is 12” from the edge and the spring is 0.75” above the surface.

F = 100 pounds.

d = 12”

h = 0.75”

x = hypotenuse

X^2 = 12^2 + 0.75^2

X = sqrt(12^2 + 0.75^2)

X = 12.023414656411049

F(lift) = 100(0.75)/12.023414656411049.

F(lift) = 6.237 lb

F (horizontal) = 100(12)/12.023414656411049

F (horizontal) = 99.805 lb

Interior angles are (90° | 86.42° | 3.576°)≈(1.571 rad | 1.508 rad | 0.06242 rad)

We have a 100 pound force being applied at 3.576°.

The vertical component is (Sin 3.576°)100 = 6.237 lb.

The horizontal component = (Cos 3.576°)100 = 99.805 lb.
 
Last edited:

Enjoying this content?

Support TFP with a donation.

Give Support
Thread Status
Hello , This thread has been inactive for over 60 days. New postings here are unlikely to be seen or responded to by other members. For better visibility, consider Starting A New Thread.