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 How much force? |
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Robse
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Dabei seit: 24.09.2011
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Guten tag :-)
First of all, sorry for my non-German. I hope that I one day will be able to meet you all in your native language, but until then I have to stay with English.
I am in the design phase of the crane boom cylinders, on my 1:14 LTM 1500-8.1... but I have a huge problem. While I can calculate how much I will be able to lift at the end of the boom, I can not find out how to include the weight of the boom it self in the calculation.
My question therefore is: How much force will the orange vector need to push with, in order to lift the boom?
Data:
The boom pivots around the red spot to the right.
Boom length: 4 meter
Boom weight: 12 kg (mass)
Distance pivot-point to point where orange vector lifts: 0,6 meter
Can anyone help me please? My head is spinning out of control over this, and I cannot agree with my self how to calculate.
Bis dan,
Robert Holsting **** www.robse.dk
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30.09.2014 15:02 |
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Kette1zu8
Foren Gott
Dabei seit: 24.08.2011
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Hi there,
you need the center of gravity! If the boom is straight and equal from the beginning to the end then you find center of gravity right in the middle. Now you have the length (i.e. 2m) and a force (12kg x 9.81) round about 120N --> 240Nm. To get the force at the cylinder pivot point you need to calculate this way: 240Nm : 0,6m = 400N (40Kg).
this calculation is for a horizontally boom and a vertical direction of Force!
__________________
Gruß, Markus
Liebherr R984 Litronic im Bau, M1:10
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30.09.2014 16:25 |
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JensR
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> how to include the weight of the boom it self in the calculation
There are detailed ways to do this, but a rough approximation is to assume the boom is weightless and half its actual weight is placed at the one end and the other half is at the other end (the pivot point and hence does not contribute to the required lifting force).
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30.09.2014 22:42 |
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Robse
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Dabei seit: 24.09.2011
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Themenstarter 
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Thank you all very much. I was afraid that the force needed was as high as 400N, but also happy that I did not go ahead and build without this taken into consideration. :-) I need to re-think my design.
Amazing how far and wide our model building takes us..
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01.10.2014 07:52 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
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Level: 27 [?]
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Themenstarter
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Hi,
Oh, great. :-)
I am aiming at the full picture, but was missing the boom as an isolated load.
I am actually working on a formula that includes the following information:
Mass, boom,
Mass, hook,
Center of (boom)gravety,
Angle between cylinder and boom,
Angle of boom vs. horizontal,
Cylinder power,
Distance, pivot point of boom -> point of lift (cylinder)
Distance, pivot point of boom -> tip of boom,
-all this to find out how strong my cylinders must be, and later make a load-chart. If I make it, I'll make an online version like the other (crane)engineering tools on my web.
I look forward to your input, and thank you very much for taking the time. :-)
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01.10.2014 10:34 |
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Dave
Eroberer
Dabei seit: 10.08.2012
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Hi Robert,
I found some draft with angles and distances, but not the calculation... tomorrow I will have more time to look after it. In the worst case I have to calculate again, but that would take a bit more time.
By the way, the time I spend on it will be a benefit for me as well 
__________________
Gruß, Davis
--------------------------------------------------------------
Wenn die Klügeren immer nachgeben, passiert nur das was die Dummen wollen 
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01.10.2014 22:36 |
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Robse
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Themenstarter
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Hi,
Thank you very much !
When I get close enough to my own pc again, I will try to show what I got so far.
Brgds, Robert
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02.10.2014 14:18 |
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Dave
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02.10.2014 20:11 |
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Dave
Eroberer
Dabei seit: 10.08.2012
Beiträge: 61
Maßstab: 1:14,5
Wohnort: Buseck
Level: 30 [?]
Erfahrungspunkte: 281.876
Nächster Level: 300.073
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At this weekend, I will try to complete the calculation with an erected boom and some weight at the hook.
__________________
Gruß, Davis
--------------------------------------------------------------
Wenn die Klügeren immer nachgeben, passiert nur das was die Dummen wollen
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02.10.2014 20:25 |
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Robse
Mitglied
Dabei seit: 24.09.2011
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Themenstarter
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Hi Dave,
Thank you VERY much!
I will sit down with my calculator and make a calculation for my setup (Need to redo a lot, I can see), and see what kind of numbers I reach., but I can already agree with you: I AM supprised: That IS a lot of force, just to get the boom up. :-)
I'll be back.. and once Again: thank you for taking the time.
Brgds,
Robert
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02.10.2014 20:25 |
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Robse
Mitglied
Dabei seit: 24.09.2011
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Themenstarter
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Hi Dave,
Two small questions, please:
The "80" on the left, distance between center of boom and lower point of cylinder is not used, right?
Where are the "7,13 degree" measured?
(Cylinder angle in ref to boom?)
Brgds,
Robert
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02.10.2014 21:05 |
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Dave
Eroberer
Dabei seit: 10.08.2012
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The '80' is a '40' 
On the bottom of the paper is a triangle of the forces/dimensions, that's what I needed to get the 7,13°. Yes, it's the angle between boom and cylinder.
__________________
Gruß, Davis
--------------------------------------------------------------
Wenn die Klügeren immer nachgeben, passiert nur das was die Dummen wollen
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02.10.2014 21:31 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
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Level: 27 [?]
Erfahrungspunkte: 153.192
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Themenstarter
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02.10.2014 21:37 |
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Robse
Mitglied
Dabei seit: 24.09.2011
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Level: 27 [?]
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Themenstarter
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Hi Again,
It looks as if I might get away with it a little easier than the example, unless I messed up that is.. :-)
This example is for when the boom is not extended, and horizonal.
The strange PI() and /180 in the SIN and COS calculation is caused by Excel. Excel reads radians as standard, everyone else uses degrees... Thus the conversion.
I'll Work on it more, and add more variables... load capacity, boom angle, extended boom etc.
Brgds,
Robert
| Dateianhang: |
 RHallin.jpg (57,50 KB, 223 mal heruntergeladen)
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Dieser Beitrag wurde 1 mal editiert, zum letzten Mal von Robse: 03.10.2014 19:10.
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03.10.2014 19:06 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 153.192
Nächster Level: 157.092
Themenstarter
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Hi again,
Well... I have got a little longer, but not quite there yet.
First I can calculate the power needed to raise the boom if it is fully horizontal, and fully extended (3,4m max length) (Worst case scenario):
Pic 1
I then use that number to calculate the available power used for hooks and load:
Pic 2
What would be nice is to merge the two... so you can input cylinder power, boom angle and level of extension, and then get "max load". To get to that, the first formula must be extended to include boom angle.. and then the two formulas must merge.
Best regards,
Robert
| Dateianhänge: |
1.jpg (57,17 KB, 176 mal heruntergeladen)
2.jpg (73,11 KB, 205 mal heruntergeladen)
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Dieser Beitrag wurde 2 mal editiert, zum letzten Mal von Robse: 09.10.2014 08:56.
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09.10.2014 08:53 |
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Dave
Eroberer
Dabei seit: 10.08.2012
Beiträge: 61
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Good morning Robert,
the same problem as I had, merging tese calculations
Sorry for not-answering in the last few days, I have to get my workshop ready until the electrcian comes on saturday to connect my new fuse box
__________________
Gruß, Davis
--------------------------------------------------------------
Wenn die Klügeren immer nachgeben, passiert nur das was die Dummen wollen
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09.10.2014 09:13 |
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