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wsuwrhr
10-14-2003, 07:15 PM
http://www.***boat.com/image_center/data/520/63boathang.jpg
I was going over some old boat pics getting everything together before I lose it again in the next move.
I thought I would put this up for anyone wondering how they would get the boat off the trailer for upcoming repairs.
What you can't see are two 8x8's in the rafters. I have 4x6's going from the concrete up to the ends of the 8x8's. I have all the boards nailed together temporarily so nothing comes apart during lifting. The boat is lifted with two yank straps and comealongs and load spreaders. I was planning on making cool billet load spreaders and using electric winches one day. I used to have lots of time on my hands.
If I remember correctly I calc'ed the
load-bearing capability of the 8x8 beam with a static load, 9 feet between supports at 4,000 pounds per beam.
Maybe some of the guru engineers here can chime in and tell me I am crazy.
Brian
[ October 14, 2003, 08:17 PM: Message edited by: wsuwrhr ]

Rexone
10-14-2003, 07:17 PM
wsuwrhr:
Maybe some of the guru engineers here can chime in and tell me I am crazy.
Brian ok you're crazy. :D

cal***boat
10-14-2003, 07:34 PM
Here ya go I have it all figured out for you.
9y-(2y+4) 7c-(8c+2) 8w-3(5w-8)
6a+2c-3(2a+3c)
3(5+4(3-y)
5i-(3=9i)
=4001 lbs
ERROR ANALYS
1.2x+7.7=2.9
1.2+7.7=(-7.7)=2.9+(-7.7)
1.2x= -4.8
1/12-1.2x=1/1.2-4.8
x= -4
3986.28 lbs
So yes you are crazy

rrrr
10-15-2003, 08:19 AM
cal***boat:
Here ya go I have it all figured out for you.
9y-(2y+4) 7c-(8c+2) 8w-3(5w-8)
6a+2c-3(2a+3c)
3(5+4(3-y)
5i-(3=9i)
=4001 lbs
ERROR ANALYS
1.2x+7.7=2.9
1.2+7.7=(-7.7)=2.9+(-7.7)
1.2x= -4.8
1/12-1.2x=1/1.2-4.8
x= -4
3986.28 lbs
So yes you are crazy You have a decimal point in the wrong place. :D :D

JetBoatRich
10-15-2003, 08:36 AM
rrrr:
cal***boat:
Here ya go I have it all figured out for you.
9y-(2y+4) 7c-(8c+2) 8w-3(5w-8)
6a+2c-3(2a+3c)
3(5+4(3-y)
5i-(3=9i)
=4001 lbs
ERROR ANALYS
1.2x+7.7=2.9
1.2+7.7=(-7.7)=2.9+(-7.7)
1.2x= -4.8
1/12-1.2x=1/1.2-4.8
x= -4
3986.28 lbs
So yes you are crazy You have a decimal point in the wrong place. :D :D I saw that, but with a few corrections we can work through this error eek! .
Half full kind of person

cal***boat
10-15-2003, 08:38 AM
rrrr:
cal***boat:
Here ya go I have it all figured out for you.
9y-(2y+4) 7c-(8c+2) 8w-3(5w-8)
6a+2c-3(2a+3c)
3(5+4(3-y)
5i-(3=9i)
=4001 lbs
ERROR ANALYS
1.2x+7.7=2.9
1.2+7.7=(-7.7)=2.9+(-7.7)
1.2x= -4.8
1/12-1.2x=1/1.2-4.8
x= -4
3986.28 lbs
So yes you are crazy You have a decimal point in the wrong place. :D :D :confused: Are you sure, I have gone over the calulations 5 time and I can't find the error. :confused:
idea Now if I had a little more information on just how the gismo is built I could get it down to the ounce. idea

wsuwrhr
10-15-2003, 03:21 PM
cal***boat:
[QUOTE] :confused: Are you sure, I have gone over the calulations 5 time and I can't find the error. :confused:
idea Now if I had a little more information on just how the gismo is built I could get it down to the ounce. idea Well geez, that is all you had to say. What else do I need to include?
Brian

sidewound
10-15-2003, 03:27 PM
You need to include the bullet variance factor along with the decimal point. Add a left handed twistie and the whole formula falls apart. Just not feasable!
Peace Man :cool:
CESAR

ROZ
10-15-2003, 03:46 PM
I think that equations needs some Pi in there...Hair Pi that is..... :D
Or as The Duke put it, reaaaaaaally hairy pie wink :D

cal***boat
10-15-2003, 03:55 PM
sidewound:
You need to include the bullet variance factor along with the decimal point. Add a left handed twistie and the whole formula falls apart. Just not feasable!
Peace Man :cool:
CESAR Damn you took the words right out of my post, I forgot all about the bullet variance because that would throw the whole thing off and make it very unstable and the weak link would be the left upright at the convergence point with the center cross section.
You could proberly install 2 vertical force limiters onto the cross section at 25" inboard from the uprights to increase it's rating by 8 percent which still might not be enough to do the job.
Try 2x=3c(4ix2c-4a)=6j)03b-2.2 and it might work, you can divide this calculation by 100 and build it from tooth picks and the divide the weight of the boat by 89.6 percent to find an equal weight to test it with.
The 14 years i spent in college really helps with this kind of stuffs and working with gizmos.
[ October 15, 2003, 06:14 PM: Message edited by: cal***boat ]

cal***boat
10-16-2003, 05:36 AM
:confused: :confused: idea :confused: :confused: