Can anyone please confirm my Pythagorus workings ?

[MtB]

 

 

Let's hope you're right or something like THIS could happen....

 

 

 

 

 

 

Actually, the sign writing is wrong, they don't fit!!

 

 

 

[David Mack]

Or this

[MtB]

To be fair, on my photo it looks like the lorry might have fitted ok but had to brake hard and began to jack-knife which tipped the trailer over like that. 

 

 

[cuthound]

45 minutes ago, Mike the Boilerman said:

To be fair, on my photo it looks like the lorry might have fitted ok but had to brake hard and began to jack-knife which tipped the trailer over like that. 

 

 

 

I thought perhaps the driver hadrealised he was too high to go under the bridge and had swerved in the vain hope he might tip over enough to get under it. ?

[WotEver]

53 minutes ago, Mike the Boilerman said:

To be fair, on my photo it looks like the lorry might have fitted ok but had to brake hard and began to jack-knife which tipped the trailer over like that. 

And in the second photo I’m guessing the driver forgot he had a double decker. 

[TheBiscuits]

So having lifted and moved your boat, how accurate were your calculated numbers?

 

(I am assuming someone measured it while it was out of the water!)

 

@Alan de Enfield obviously

 

Edited October 25, 2019 by TheBiscuits

[Alan de Enfield]

Weeeeeelllllllll - its all a bit complicated as I measured the boat draft at the widest part (therefore using more 'string') this gave a draft of about 6 feet.

 

The actual deepest part of the boat is at the stern where the boat is only 9.8 feet beam - I suppose I should have tried the Pythagorus method again whilst out of the water.

 

With the boat 'level' on the ground - with 8" of planks beneath the keel at the stern,  there were 24" of boards ( 4x 8" Railway sleeper offcuts) at the bow, so, the draft at the keel is about 16" more at the stern.

 

Measuring (at the stern) from the ground to the 'water tide mark' and subtracting the 8" of wood gave a draft of 57" ( 4' 9")

 

Basically - the string didn't work being 1' 3" 'out' but measured at a different point.

 

 

Picture with boat waterline 'horizontal, at 8" off the ground at the stern and over 24" of chocks 10 feet back from the start of the bow uplift.

Edited October 25, 2019 by Alan de Enfield

[rusty69]

So....the moral of the story is never trust Pie fag or us to measure your hippos arse at noon.

[Mac of Cygnet]

Why do you think your triangle is right-angled?  Pythagoras only works with those.

[Alan de Enfield]

1 minute ago, Mac of Cygnet said:

Why do you think your triangle is right-angled?  Pythagoras only works with those.

By taking 1/2 of the overall length, and 1/2 of the boat beam it gives a right-angled triangle

 

I took a length of rope tied onto the railings, under the keel, up the other side and tied onto the railings.

I hung upside down gripping with my toes and tied a cable tie around the rope (on both sides) to mark the water line.

Slid the rope back from under the boat and measured the length between the cable ties as 5.35 mts.

 

Halved the rope measurement to get the 'hypotenuse ' (2.675m)

 

Measure the beam of the boat (as best as I could at the waterline (not easy !!!) and measured about 3.75m.

Halved this to get the 'base' of the triangle  giving 1.875m

 

Hypotenuse C² = Height B² + Base line A² 

 

So wanting to find the height B² = C² - A²

 

So √ B = √ (C² - A²)

[Machpoint005]

39 minutes ago, Alan de Enfield said:

By taking 1/2 of the overall length, and 1/2 of the boat beam it gives a right-angled triangle

 

I took a length of rope tied onto the railings, under the keel, up the other side and tied onto the railings.

I hung upside down gripping with my toes and tied a cable tie around the rope (on both sides) to mark the water line.

Slid the rope back from under the boat and measured the length between the cable ties as 5.35 mts.

 

Halved the rope measurement to get the 'hypotenuse ' (2.675m)

 

Measure the beam of the boat (as best as I could at the waterline (not easy !!!) and measured about 3.75m.

Halved this to get the 'base' of the triangle  giving 1.875m

 

Hypotenuse C² = Height B² + Base line A² 

 

So wanting to find the height B² = C² - A²

 

So √ B = √ (C² - A²)

 

I see you have "come over" to proper metric units to do a sensible calculation! Was the 144-times-table too much to bear?