Lifters needed for two sea kayaks

[kitmann]

We purchased a mid-90s Hawk a couple years ago and I have been customizing. We want to carry our two 17' fiberglass sea kayaks on our annual sojourn into Baja. The boats with a few things stored in them weigh about 75 lbs each, plus weight of long Yakima bars, so something a little south of 200 lbs overhead. I know people talk about removing boats before lifting, but that just isn't practical for quick overnight traveling stops.

Following the previous threads regarding lifters, I started with two 60 lb on the front and two 40 lb on the back. Short story, the 40's on the back were not enough. Now have 60's all around. Still takes both of us to lift but it can be done.

When we do not have the kayaks, I simply remove one of the lifters from front and back and the remaining single lifter is a nice assist for unloaded roof.

It is obvious to me now, but just a reminder to others that are considering this, remember that the lifters have very little effect at the beginning when the roof is closed. All the force is going laterally. It's only after getting the roof up a little ways and they begin to be pointed up that they kick in. So that first few inches is the hard part.

I've seen trigonometry in these posts, so I know there are some science minded folks out there. My physics question: as the lifter goes from nearly horizontal, with a very small upward force vector, to the fully open position with a high angle and much greater upward force vector, does the upward force change arithmetically, geometrically or logarithmically as the angle steepens? Is it a straight line relationship or a curve?

Also, I now have two 40 lb lifters for sale, PM me if interested.

 

[MarkBC]

...
I've seen trigonometry in these posts, so I know there are some science minded folks out there. My physics question: as the lifter goes from nearly horizontal, with a very small upward force vector, to the fully open position with a high angle and much greater upward force vector, does the upward force change arithmetically, geometrically or logarithmically as the angle steepens? Is it a straight line relationship or a curve?
...

I believe the vertical force is proportional to the cosine of the angle between the strut and vertical.

So, when the strut is almost horizontal (when the roof is down), almost 90° with vertical, the vertical force is almost zero, because cos(90°) = 0. So none of that 60lb would be in the vertical direction.
And as the roof lifts and the strut angle becomes more vertical then at the extreme if the strut was vertical: cos(0°) = 1, so the vertical force would be 60lb. (but the strut can't ever be vertical, since the roof-top bracket isn't vertically above the lower bracket)
And in between those extremes the force varies as the cosine of the angle...so I guess you could call that dependence "geometric", or better, "sinusoidal" with angle.

And the same thing when lowering, of course, which is why I have to pull hard to get it to start down.

All this assumes that the struts provide a constant force, e.g., 60 lb, at all extensions. Do they?

 

[MarkBC]

If you want to know the vertical force as a function of the vertical elevation -- without having to measure the angle:

Vertical Force = 60 lb x [(y/a)/√((y/a)² + 1)] (notice the square root sign -- √ -- and of what you're taking the root)

"a" is the horizontal distance between the two ends of the strut -- which is a constant, fixed number.
"y" is the vertical distance between the two ends of the strut, which increases as the roof goes up.
The length of the black line is the current length of the strut, which increases as the roof lifts...but you don't need to measure that if using this formula.

You could just measure the lengths of the red line and the black line at any elevation and the ratio of red/black gives you the fraction of the 60 lbs that's vertical. But using the formula above you only have to measure one changing distance -- the vertical.

Don't get confused between the triangles in the my two posts: The first one is a combination of force vectors, in which the hypotenuse (the black arrow) is a constant 60 lb and both of the red arrows vary in magnitude, and the second one is actual distances, in which only "a" is constant.

 

[buckland]

Retired Math teacher here with a 2011 Eagle... Started with what came from factory... I realized I would like it easier to lift... So I bought the speaker stand crank. It was fantastic... and still is. It does exactly what you need... getting the lift started. I found though that after a Yakima set up with bars and a rack for some gas cans... the crank was working but complaining too. So I started into the long process of putting on the lift struts on an Eagle which is different than others as to the struts you will need to order... ASK FWC. I did add the struts and still found with the weight up there, the start was a bit of push (thinking if I had had a beer or three first... is this a good idea? So I carry the crank s well... very light weight and a very helpful start to the lift..... why hurt yourself? Hope this helps... works for me.
PS I love math... geometrically.

 

[jha6av8r]

I have 100# internal struts and carry a hand crank.

 

[JaSAn]

+2 on the speaker stand. I got to 65 with a good back by not taking unnecessary chances.

...
All this assumes that the struts provide a constant force, e.g., 60 lb, at all extensions. Do they?

No. A strut is a gas (nitrogen) shock. 60 lbf is the maximum force at max compression (at some defined temperature; remember your PVT diagrams?). Force at max extension will be based on the change in volume. Should be linear.

jim

 

[MarkBC]

A strut is a gas (nitrogen) shock. 60 lbf is the maximum force at max compression (at some defined temperature; remember your PVT diagrams?). Force at max extension will be based on the change in volume. Should be linear.

jim

Hmm..makes sense...but also means that the force rating of a strut is really only meaningful in comparison to other similar-sized struts. That is, a strut with a longer or shorter stroke would have a more or less loss in force as it extended. Good to know.
Also handy that the increasing vertical lift caused by the change in angle goes the opposite direction of the decreasing gas force. Otherwise they wouldn't be much good for lifting.

I see, as described here, I guess. That's really interesting!

So those equations that I posted above would have to be modified by dividing them by the ratio of (extended volume/compressed volume)...if someone really wanted to know how the roof-raising force varies as the roof moves up or down.

 

[JaSAn]

... "So those equations that I posted above would have to be modified by dividing them by the ratio of (extended volume/compressed volume)...if someone really wanted to know how the roof-raising force varies as the roof moves up or down."


You also need to know the force at the end of travel (max extension).
So your equation would read:

Fy = Fmax CosØ - ((Fmax - Fmin) x') CosØ

where
Fmax is force at full compression of the strut
Fmin is force at full extension
Fy is the force pushing up
x' is percent strut is extended
Ø angle between shock and vertical

Change in force should be linear.

Think that is right. Brain gone to mush, taking a break from doing taxes tonight.

jim

 

[MarkBC]

OK, I think I agree with that (after the edits... )

Fy isn't really linear with Ø or with y...but not too far off over some of the range.

I wonder if "Fmin" is a listed spec for these struts we're using...?..I bought mine a few years ago and don't remember. Seems like that's a parameter that could mater to people if it's a lot different than Fmax

 

[kitmann]

Thanks for the technical discussion MarkBC and JaSAn.
Considering Mark's first comment, that the force is proportional to the cosine of the angle; a cosine 'curve' would be very nearly straight between the closed position, where the lifter is at a few degrees angle, to the open position where it's maybe 60 or 70 degrees. Ergo, change in force is (essentially) linear. The further consideration that force change due to piston position is also linear (although in the opposite direction?), the whole thing would be linear.
So why does it seem like the first few inches are so much harder than the rest?

 

[MarkBC]

..... a cosine 'curve' would be very nearly straight between the closed position, where the lifter is at a few degrees angle, to the open position where it's maybe 60 or 70 degrees...

So why does it seem like the first few inches are so much harder than the rest?

Yeah, you're right -- the cosine curve is not far from linear in that part of the curve (even if not actually linear).

"Why does it seem...?" Couldn't it be that when the roof is down and you have almost NO help from the strut, because of the angle, that it just seems like a huge weight. (like, "Why isn't this thing working?!" ) And when you manage to lift it a little and change the angle then you get just a little help and then a little more help -- and get some inertia going, too -- that it then seems so much easier?

Maybe your perception is not comparing the (nearly) linear arithmetic increase in "help", instead you're comparing the ratio of help you soon get to the initial help you get (none). And with "none" in the denominator...
Perception -- for sure -- is nothing like linear!