# Can I load my boat without a winch?

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• oldGuy53
In summary, if the boat is 10' long and the weight is evenly distributed over its length, it would require 20 lbs. of pressure at the "heaviest" it will seem to move it from vertical to horizontal.

#### oldGuy53

I have a boat that weighs about 100 lbs. and is 10’ long. I’d like to load it on my pickup rack in the manner shown in this (and many others) video, but without having to purchase and mount a winch.

I’m fairly certain I can get the boat upright. The portion I don’t know how to calculate is once it’s standing up. Pulling the transom away from the truck should be fairly easy but from there to the upside down horizontal position eludes me.

The rack is 7’ from the ground. The transom is about 2’ tall. So when the boat is upright, the balance point will be close to the rack.

I suppose I could attach a block and tackle to the front rack and pull the bow with it.

How would I calculate the maximum effort required.

Thanks for reading. I look forward learning how to answer this question.

Ken

Are you saying its going to be pivoting about its center of mass? This might be trickier than I think, but I suspect initially when the boat is near vertical, since there isn't going to be a lot a friction on the roller, you are basically going to have to support nearly the full weight of the boat. That force should diminish as ##W \cos \theta \to 0## as ## \theta \to \frac{\pi}{2} ##

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There are side-load hand-lift assist setups that folks have used. Have you looked into that option?

Also, this is my wife's kayak that she is able to load up onto the top of my Jeep Grand Cherokee by hand using leaning leverage from the back of the vehicle. The roof is probably 6' tall and my 60 y/o wife is about 5' tall. Her 10' kayak weighs about 60 pounds (so I can lift it up and onto the Jeep roof rack no problem by myself from the side, but she often goes kayaking by herself).

Can you post pics of your boat and vehicle? Are you comfortable lifting avout 60 pounds to leverage your 100 pound boat up onto the back of your pickup rack (using good lifting technique)?

Thanks to all.

Yes, I can handle 60 lbs, if it’s not awkward.

Berkeman, I have looked at the side lift options. Just hoping to keep it very simple. They also require manually flipping the boat over.

Erobz, at vertical, the lines tied between the rear rack and the transom should be supporting the weight, I think.

After vertical, the commercial units use a winch to pull the forward part of the boat onto the rack, causing it to tilt to horizontal. I’m hoping to manually swing and lift the aft of the boat. Moving the transom from the 6 o’clock position to the 3 o’clock position is the question. How heavy will it be, realizing the it is secured to the rack, and pivoting at close to the mid point?

We’ll, I’m 69, but I’m thinking it’s doable. Probably easiest to just try it.

Thanks.

What is the fence thing at the transom? Is it fixtured to the ground? Do you need one where you unload? Generally I would not try that rig without a winch.
But then I'm seventy...

Welcome!
Does your rack like these ones?

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The problem is that the vehicle is 7 ft high and the boat is only 10 ft long. That is a difficult lift for a human body.
oldGuy53 said:
I suppose I could attach a block and tackle to the front rack and pull the bow with it.
Not the bow. Pull with one rope that splits into a bridle, attached to the rowlocks inside the boat. That way the boat will first rise, and not push so hard on the back of the vehicle. The rear bar will be a roller that supports the lifting rope and gunwales as it stands and then rises.

I would start with the stern to the vehicle, attach the bridle ends to the rowlocks, then the boat would stand on it's transom before it rises up onto the roof racks in one long movement.

hutchphd and Lnewqban
Okay. I think I didn’t express the question well.

Let’s forget about the boat for a minute.

There is a beam that is 10’ long and weighs 100 lbs. It is vertical and attached to a pivot point. The top end of the beam is A. The bottom end of the beam is B. The pivot connection is C. C is 6’ from B and 4’ from A. The weight is evenly distributed over the length.

Applying pressure ONLY to point B, what is the maximum effort required to move the beam from vertical to horizontal? Or, what is the “heaviest” it will seem?

What is the equation?
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My thought is that at horizontal, the apparent weight of B will be 20 lbs. - 60-40. Is this valid?

At every degree of tilt between vertical and horizontal, the “weight” will be different, correct? Will it be more than 20 lbs at any point? If so, how is that calculated?

Thanks to all.

Lnewqban said:
Welcome!
Does your rack like these ones?

Like the first one. But my boat is much smaller.

Lnewqban
Baluncore said:
The problem is that the vehicle is 7 ft high and the boat is only 10 ft long. That is a difficult lift for a human body.

Not the bow. Pull with one rope that splits into a bridle, attached to the rowlocks inside the boat. That way the boat will first rise, and not push so hard on the back of the vehicle. The rear bar will be a roller that supports the lifting rope and gunwales as it stands and then rises.

I would start with the stern to the vehicle, attach the bridle ends to the rowlocks, then the boat would stand on it's transom before it rises up onto the roof racks in one long movement.
I think, but don’t know, that I don’t need a lifting rope at all. I hope I don’t. Thus the question here about the effort required.

To stand the boat up on the transom will take less than half the 100 pounds. But to then lift the hull to the point where you can begin to slide it forward, you will need to lift the entire weight of 100 pounds.

There is a safety issue here. On the day you cannot quite lift it all the way, you will have the boat fall back on you from above. At that moment, you will be at your most vulnerable, and will probably be physically injured. You will also notice the mental effort necessary before lifting 100 pounds up 7 feet, becomes more difficult to attain.

A winch and bridle operates to protect the operator in several ways. It keeps you out of harms way, and takes the weight when you need to stop. By selecting the bridle attachment points, it will also control the orientation of the hull during the lift.

berkeman and Lnewqban
oldGuy53 said:
Okay. I think I didn’t express the question well.

Let’s forget about the boat for a minute.

There is a beam that is 10’ long and weighs 100 lbs. It is vertical and attached to a pivot point. The top end of the beam is A. The bottom end of the beam is B. The pivot connection is C. C is 6’ from B and 4’ from A. The weight is evenly distributed over the length.

Applying pressure ONLY to point B, what is the maximum effort required to move the beam from vertical to horizontal? Or, what is the “heaviest” it will seem?

What is the equation?
========
My thought is that at horizontal, the apparent weight of B will be 20 lbs. - 60-40. Is this valid?

At every degree of tilt between vertical and horizontal, the “weight” will be different, correct? Will it be more than 20 lbs at any point? If so, how is that calculated?

Thanks to all.
If the beam is fixed to the pivot ( like a sea-saw ) and you lift vertically at point ## \boldsymbol B ## then the force you will need to apply for the given dimensions at minimum is:

$$\frac{W}{6}$$

It is independent of the angle, because you are not required to support the vertical component of the beam weight in this case.

In general ( applying a vertical force ):

let ## x ## be the distance from ## P \to A ## and ## l ## be the total length of a beam with uniformly distributed mass per unit length:

For ## 0 \leq x \leq \frac{l}{2} ##

$$\frac{1}{2}W \left(1 - \frac{x}{l-x} \right)$$

Note: The actual force you need to apply will be dependent on how rapidly you wish to change its angular velocity. The solution I provided is for (what I believe to be known as ) "quasistatic equilibrium"...meaning its moving, just barely.

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oldGuy53 said:
We’ll, I’m 69, but I’m thinking it’s doable. Probably easiest to just try it.
Yes it probably is (although you might want to have a friend standing by to help if you can't manage).

I'm only 58 and my car's roof is only about 5' high but I'd be happy loading my 130lb boat in this way.

Oh wait a minute...
oldGuy53 said:
The rack is 7’ from the ground.
6' I'd be happy with (I'm 5'10) but at 7' there's different muscles involved and a lot less control. How well this works depends on the design and manufacture of the pivoting frame, I'd try it the first time with somebody actively helping rather than just standing by.

Unless you do something clever this is not a good idea. Suppose we start with the transom on the ground, the boat vertical leaning on the roof rack, keel outwards. The first thing you will need to do is lift the boat straight up (100lbs) for a foot or more before attempting to go more horizontal.
Unless a helicopter is involved your pivoting at point C is not a representation of reality IMHO.

erobz
Another part I didn’t explain well: There are two ropes attached between the transom and the rack. As the boat is rotated to vertical, they are holding the full weight, creating the pivot point. I would not be lifting the full weight at any point.

Erobz, thanks for the formula. I don’t understand it yet, but I’m working on it.

erobz
oldGuy53 said:
There are two ropes attached between the transom and the rack. As the boat is rotated to vertical, they are holding the full weight, creating the pivot point.
Those two ropes change the game, so long as they are attached to the top of the transom and do not stretch.

pbuk
oldGuy53 said:
There are two ropes attached between the transom and the rack. As the boat is rotated to vertical, they are holding the full weight, creating the pivot point.
I am not seeing this, absent a winch. How do these ropes engage to support the full weight of the boat?

There are two lifting handles on the aft side of the transom. Almost all small boats have these. The ropes are just long enough to attach to these handles and to the rack. As the bow is lifted, the transom comes off the ground, the the weight supported by the ropes. Once the boat is vertical, these ropes effectively create the pivot point. You can see this happening in the videos.

Baluncore said:
Those two ropes change the game, so long as they are attached to the top of the transom and do not stretch.
Possibly, but with the dimensions provided I'm not so sure.

The boat is 10' long and the rack is 7' off the ground. That means that as you lift the transom up and out, only 3' of the bow, which is the lightest part of the boat, is going to be supported by the pivot point at the rack - at best 20% of the weight of the boat, so you are still going to be lifting over 80 lbs above head height.

pbuk said:
Possibly, but with the dimensions provided I'm not so sure.

The boat is 10' long and the rack is 7' off the ground. That means that as you lift the transom up and out, only 3' of the bow, which is the lightest part of the boat, is going to be supported by the pivot point at the rack - at best 20% of the weight of the boat, so you are still going to be lifting over 80 lbs above head height.
Agreed, his boat is not a beam with uniform mass distribution as he asked to model it as. The actual CoM may be more like 6 ft from the tip of the bow. That obviously depend on the shape of the hull.

oldGuy53 said:
Like the first one. But my boat is much smaller.
Any way to show us a couple of pictures of the rack?

oldGuy53 said:
There are two lifting handles on the aft side of the transom.

I see, thanks.
So if you lift the bow by hand to vertical (which is probably easiest for control) that will be <50lbs. The next move is the tricky one then

Truly, the CoM coordinate and the boat need to be considered in 2 dimensions. The CoM is likely closer to the floor of the boat, which means it is offset from the pivot maybe a foot or more. This increases the torque, and the point application of the force on the extended body also becomes relevant.

Basically, you can disregard that simple 1D beam model I provided earlier.

hutchphd said:
So if you lift the bow by hand to vertical (which is probably easiest for control) that will be <50lbs. The next move is the tricky one then
But that next move is not a vertical lift. It is a rotation about the rear roof bar, while the whole weight is initially supported by the ropes. You hold the transom, then walk backwards away from the vehicle while half the hull weight is being transferred from the ropes onto the rear roof bar. You must support half the weight just before sliding the hull forwards.

How close to prow is the roof bar? This could be pretty unstable and tippy if it slides sideways. It will not be full weight but unwieldy and then half weight at an awkward height. Devil in the details .

hutchphd said:
How close to prow is the roof bar?
10 ft long boat - 7 ft high bar = 3 ft.
Gunwales are then probably more than 2 ft apart at that pointy end.

Yeah that maybe seems do-able although you probably want the boat a little past vertical when you switch from lifting the bow up and finishing with the stern lift. Remember you have switch from lifting the bow to lifting the stern while the boat is dangling.