How to weigh a car topped boat?

  • I
  • Thread starter HRG
  • Start date
  • Tags
    Boat Car
In summary, the OP is planning to build a 12 foot jon boat that will be car topped on their van. The rated loading on the roof racks is 165 lbs. He plans to weigh the finished boat by lifting the front and back of the boat using two bathroom scales and calculating the weight using the equation W=H+F.
  • #1
HRG
32
8
I'm planning to build a 12 foot jon boat that will be car topped on our van. The rated loading on the roof racks is 165 lbs.

To weigh the finished boat, I plan to do the following:
  1. Position the boat upside down in our carport.
  2. Stand on a bathroom scale and lift the front of the boat. Write down the weight (minus my body weight).
  3. Stand on a bathroom scale and lift the back of the boat. Write down the weight (minus my body weight).
  4. How can I calculate the weight of the boat using those two weight values?
Thanks.
 
Physics news on Phys.org
  • #2
HRG said:
How can I calculate the weight of the boat using those two weight values?
You can't. That doesn't tell you the weight of the boat.

Can you rest it on two bathroom scales?
 
  • #3
Vanadium 50 said:
You can't. That doesn't tell you the weight of the boat.
Why not (if the boat is horizontal for both readings)?

But why not just sit the boat (balanced) directly on the scales, using blocks (bricks - everyone has a few bricks). OR weigh each end and add the weights.
 
  • Like
Likes russ_watters, 256bits and Ibix
  • #4
Vanadium 50 said:
That doesn't tell you the weight of the boat.
Say the boat has length ##l## and weight ##w## and its center of gravity is ##x## from the line on which its stern rests. Lift the prow a couple of millimeters. The weighing scale will read ##W_{prow}=H+F_{prow}## where ##H## is @HRG's weight and ##F_{prow}## is the force applied to the boat which must satisfy ##F_{prow}l=wx##. Then go to the stern and lift the boat a couple of millimeters again. The weighing scale will read ##W_{stern}=H+F_{stern}## where ##H## means the same as before and ##F_{stern}## satisfies ##F_{stern}l=w(l-x)##. Then it's just simultaneous equations, surely? I mean, it's important that he lifts the boat as little as possible so he doesn't need to measure angles and do trigonometry but even that wouldn't be insurmountable.
 
  • Like
Likes russ_watters
  • #5
The total force upwards has to equal the weight. Measuring the two downward forces (weights) is valid if the boat's angle doesn't change. You can test the effect of height / angle by varying the number of bricks. Not a significant problem when you consider the additional forces acting on a roof rack as you drive along with a boat up there (wind, corners and bumps!). It would be a pity to knacker the car roof.
I suggest the OP looks at images of other users of his class of boat and see how they work on rooftops.
 
  • #6
Ibix said:
its center of gravity is x from the line on which its stern rests
Ah...but how do you know that?

sophiecentaur said:
Why not (if the boat is horizontal for both readings)?
What makes the boat horizontal on each weighing?
 
  • #7
Vanadium 50 said:
What makes the boat horizontal on each weighing?
That would be the guy doing the measurement.
 
  • Like
Likes russ_watters
  • #8
Anyway, it's not that the boat can't be at an angle for the two measurement approach. It's that the angle can't change for the two measurements.
 
  • Like
Likes 256bits
  • #9
Vanadium 50 said:
Ah...but how do you know that?
How do I know what? ##x##? Because I've got two simultaneous equations and two unknowns, ##x## and ##w##. Everything else is directly measurable.
 
  • Like
Likes sophiecentaur
  • #10
sophiecentaur said:
Why not (if the boat is horizontal for both readings)?

But why not just sit the boat (balanced) directly on the scales, using blocks (bricks - everyone has a few bricks). OR weigh each end and add the weights.
He could also wrap a rope around the centre of mass of the boat, loop the rope around a 'frictionless' pulley, and note the decrease in the bathroom scale from his body weight to the new measured weight as he lifts it off the ground.
 
  • #11
The approach in this thread has been largely from a mathematical point of view when what is needed is a Engineering approach. Going over a bad bump or pothole in the road can increase the load on the rack by say 50%. OK, the rating of the rack may have included this sort of thing but it's not something I would want to 'experiment' with with my own car.
Measuring the weight of the boat by adding the contributions to its weight at each end is perfectly adequate for any relevant decision.
IMO it would be essential to look at a lot of images of cars with boats on their roofs and to post on boating forums to get much more valuable information than you will get from here - although our general chat is always useful. Post the question on the right forum and you can expect helpful personal information - plus advice on how to get the boat up there and down again without hurting yourself, the car or the boat.
 
  • Like
Likes Vanadium 50 and DaveE
  • #12
sophiecentaur said:
The approach in this thread has been largely from a mathematical point of view
I agree, and I'd go one step more - it's being treated as a homework problem, where the consequences are that you might get a point off, not as a real-life problem with safety concerns. Ever been on the road where a car ahead of you loses a canoe? "Lieutenant, bring me my brown pants".

Sure, you can get an answer with balancing forces, torques, (stretchless ropes, frictionless planes, etc.) But I would argue that if you do that, you need to propagate the uncertainties. There's a big difference between 150 ± 5 pounds and 120 ± 60 pounds.
 
  • Like
Likes sophiecentaur
  • #13
Vanadium 50 said:
There's a big difference between 150 ± 5 pounds and 120 ± 60 pounds.
Yeah. Much safer to talk to the man who has made one and used it for 20 years. Or a manufacturer with a guarantee.
Edit: I was yakking on about Engineering and never spotted what could be a flaw about the OP. If the limit for the roof rack is 165lbs then surely the boat should be designed and built with that figure in mind. If the boat turns out to be over weight then a new rack and a new van may be required.
Edit Edit: I just read on Google that a typical 12 ft Jon boat weighs about 120lbs. But a home made boat can be well over weight if every board and rib is chosen thicker than the design calls for (to be on the safe side). I remember looking over an Enterprise sailing dinghy that had been made by a very competent joiner but it was waayyy over weight (could hardly move it) and I gave it a miss, choosing one that was only just above the bare minimum.
 
Last edited:
  • #14
Vanadium 50 said:
You can't. That doesn't tell you the weight of the boat.
Can you rest it on two bathroom scales?

Unfortunately, no. We only have one bathroom scale.

I'm thinking of the problem like this:
Assume a 12 foot long piece of 10" x 10" lumber weighs 100 lbs (hypothetically).
If both ends are resting on saw horses, then each horse is supporting 50 lbs. So if I took the place of one saw horse and stood on a scale, the scale would read my weight + 50 lbs.

The boat is different since the weight is not distributed equally throughout its 12 foot length. But using the principle above, there must be a scientific way to calculate the weight of the boat by getting the weight at both ends.

Hope someone knows how to do the calculation.

And yes, the boat will be horizontal when weighing it at both ends.
Thanks!
 
  • #15
sophiecentaur said:
But why not just sit the boat (balanced) directly on the scales, using blocks (bricks - everyone has a few bricks).
Yes, that's an idea. Since the jon boat will have a flat bottom, I could put the boat right side up on the scale and balance the boat on the scale. Originally I was thinking how hard it would be to set the boat on it's angled transom, pointing up 12 feet high, to weigh the boat. Hence my weighing on both sides idea.

In fact I will set the bottom of the boat on the scale and balance it to confirm the "weight calculation by weighing both ends" as described in my original post.

But I still hope someone knows a formula for calculating the weight of the boat by weighing it at both ends. I'm really curious now. ... Maybe it's as simple as just adding the two weight values.

Thanks!
 
Last edited:
  • #16
sophiecentaur said:
Edit Edit: I just read on Google that a typical 12 ft Jon boat weighs about 120lbs.

Precisely what I read prior to my decision to build a car topable jon boat. I will be aiming to minimize the weight of the boat to 120 lbs.

FWIW, about 30 years ago my brothers and I built an 11 foot jon boat and car topped it. But we never weighed the boat. Back then my car was a Chevy 4 door sedan and I put add on roof racks. Much less weight capacity than the roof racks on our current Honda Odyssey van with factory roof racks. Never had a problem. My brothers also had their own roof racks for their cars to use the boat with their friends. Never had a problem. Being young and reckless, we just "assumed" that the add on roof racks were sufficient to car top the boat.
 
Last edited:
  • #17
HRG said:
But I still hope someone knows a formula for calculating the weight of the boat by weighing it at both ends.
Yes, you can just add the two weights. But, only if the angle of the boat (w.r.t. gravity) is the same during each measurement.
 
  • Like
Likes tech99 and Lnewqban
  • #18
HRG said:
We only have one bathroom scale.
They are $7.99 at Target.
 
  • Like
Likes Bystander
  • #19
sophiecentaur said:
Going over a bad bump or pothole in the road can increase the load on the rack by say 50%. OK, the rating of the rack may have included this sort of thing but it's not something I would want to 'experiment' with with my own car.
Plus the aero force at highway speeds (might be a lift force!), plus transverse wind.
 
  • #20
HRG said:
Unfortunately, no. We only have one bathroom scale.

DaveE said:
Yes, you can just add the two weights. But, only if the angle of the boat (w.r.t. gravity) is the same during each measurement.
We are technically over egging the weighing problem. School Physics tells us about net forces and the geometry tells us about the Cos function, which changes very little for small angles. If @HRG is bothered about accuracy then, as I have already mentioned, you can see how the weight at one end changes as you lift the end by a few kms. It won't be much so don't worry.

Another issue that could give less stress on the roof and the rack would be to identify where the centre of mass of the boat is (put a piece of timber underneath and see the balance position) and put that about half way between the bars. It won't help with windage or stopping the boat wanting to fly off but it will spread the load on bumps evenly. If the boat is riding hull-up then it will probably be aerodynamic enough - as long as you don't try 100mph!

Just read around as much as you can. This link has several relevant opinions. (like tying down really well)
 
  • #21
sophiecentaur said:
lift the end by a few kms
:oldeek:
 
  • Like
Likes Delta2
  • #22
Ibix said:
:oldeek:
Don't you just love spellcheckers? :biggrin: I won't re-correct it; see who else spots it.
 
  • Like
Likes Ibix
  • #23
"kms" = kilo-milli-seconds of arc. Seems reasonable to me... :wink:
 
  • #24
berkeman said:
"kms" = kilo-milli-seconds of arc. Seems reasonable to me... :wink:
you are far too tolerant of my rubbish tryping.
 
  • Haha
Likes berkeman
  • #25
HRG said:
I'm planning to build a 12 foot jon boat that will be car topped on our van. The rated loading on the roof racks is 165 lbs.

To weigh the finished boat, I plan to do the following:
  1. Position the boat upside down in our carport.
  2. Stand on a bathroom scale and lift the front of the boat. Write down the weight (minus my body weight).
  3. Stand on a bathroom scale and lift the back of the boat. Write down the weight (minus my body weight).
  4. How can I calculate the weight of the boat using those two weight values?
Thanks.
I wonder if it would be easier to use a rope harness and pulley to raise the boat- if it weighs less than you, you can stand on a scale while doing this and compare the two readings, the difference is the boat weight.
 
  • #26
Andy Resnick said:
I wonder if it would be easier to use a rope harness and pulley to raise the boat- if it weighs less than you, you can stand on a scale while doing this and compare the two readings, the difference is the boat weight.
I can't visualize what you mean.

But in analyzing this problem more, I've concluded that lifting the boat on both ends per my original post and just adding the weights will result in the weight of the boat.

My logic for that is:
If a 10" x 10" x 12' lumber weighs 100 lbs and is set on two sawhorses at both ends, then each sawhorse will be supporting 50 lbs. If I replace one sawhorse by me standing on a scale, then the scale will read 50 lbs + my weight. This is pretty straight forward.

So even if the boat's weight is not evenly distributed (transom end weighs more than bow end), just adding the two weights will equal the total weight of the boat.

Finding the balance point as one poster suggested will allow placing the boat on the roof racks so that each rack will share an equal amount of the weight.

Thanks.
 
  • Like
Likes russ_watters
  • #27
HRG said:
If a 10" x 10" x 12' lumber weighs 100 lbs and is set on two sawhorses at both ends, then each sawhorse will be supporting 50 lbs. If I replace one sawhorse by me standing on a scale, then the scale will read 50 lbs + my weight. This is pretty straight forward.
Quite correct. Though a 10 inch by 10 inch beam that is twelve feet long should be more like 250 to 350 pounds. A beam that size made of balsa would be about 60 or 70 pounds by my calculations.

Back of the envelope: Specific gravity about 0.4 to 0.7. Times 5/6 times 5/6 times 12 times 62.5 pounds per cubic foot.

HRG said:
So even if the boat's weight is not evenly distributed (transom end weighs more than bow end), just adding the two weights will equal the total weight of the boat.
Yes. The two support forces from the sawhorses must match the two support forces from you, as long as you are lifting in the same place as the sawhorses on a boat that remains in the same orientation. So the sum of the two forces from you must match the sum of the two forces from the sawhorses. That sum is, of course, equal to the weight of the boat.

One assumes that you will try to lift straight up without applying a torque. Doing otherwise could throw things off a bit.
 
Last edited:
  • #28
You will get an accurate enough result by measuring the weight at both ends and adding them, however you must make sure that you measure the weight at each end at exactly the same point as it is supported, with the boat at exactly the same angle. This is best achieved by supporting the boat on two 4x2" timbers (which I believe in the US are referred to as 2x4s) with an additional support the same height as the scale -

By the time you have arranged all of this it is probably easier just to manoeuvre the whole boat onto the scale, although it may not be easy to read under the hull of the boat.

I have measured a racing dinghy the 'end by end' way and got results within 1kg of that measured by an official class measurer.
 
  • Like
Likes sophiecentaur
  • #29
sophiecentaur said:
Going over a bad bump or pothole in the road can increase the load on the rack by say 50%. OK, the rating of the rack may have included this sort of thing
Yes, maximum load ratings for roof racks (and for car roofs, which are often less then for roof racks) are dynamic load ratings i.e. take into account normal driving conditions. In practice, with careful driving, you can carry a much larger load than the rating without damage to the car.
 
  • #30
pbuk said:
This is best achieved by supporting the boat on two 4x2" timbers (which I believe in the US are referred to as 2x4s) with an additional support the same height as the scale.
Yes, that would be the "saw horse" approach. A 2 by 4 (nominal dimensions -- actual is normally about 1 and 3/4 by 3 and 5/8) could be used as the top of a saw horse.
 
  • #31
jbriggs444 said:
Yes, that would be the "saw horse" approach. A 2 by 4 (nominal dimensions -- actual is normally about 1 and 3/4 by 3 and 5/8) could be used as the top of a saw horse.
I think using saw horses would be both less convenient and less accurate. Put the scale on the ground and the 4x2 directly on top.
 
  • #32
HRG said:
I can't visualize what you mean.

I am suggesting you rig up something like this to lift the boat:

1626133960409.png
 
  • #33
I bet the OP is sooo glad he came to PF for this question. :rolleyes:
 
  • Like
  • Haha
Likes pbuk, sophiecentaur, DaveE and 1 other person
  • #34
jbriggs444 said:
Yes, that would be the "saw horse" approach. A 2 by 4 (nominal dimensions -- actual is normally about 1 and 3/4 by 3 and 5/8) could be used as the top of a saw horse.
Yes, I will use sawhorses when I weigh the boat.

The boat will be slightly less than 4 feet wide. I'll have two sawhorses with 5 feet long 2x4's screwed to the tops which I'll be using to build the boat on. So they will already be available. I'll just set the ends of the boat on the sawhorses, then put a bathroom scale below a horse and barely lift that end of the boat off the horse while standing on the scale. Ditto for the other end. The measured weight will be way close enough to verify the loading on the roof racks.
 
  • #35
Andy Resnick said:
I am suggesting you rig up something like this to lift the boat:

View attachment 285872
Wow, way more complex than what I have in mind. But I appreciate your contribution to my question.
Thanks.
 

Similar threads

Replies
9
Views
2K
Replies
28
Views
1K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
4
Views
4K
Replies
7
Views
16K
Replies
19
Views
11K
Back
Top