Recent content by Ted Farkas

Thanks for your help, No problems, once I understand the rudimentary design i.e what's happening , we then moved forward an get it signed off before manufacture. regards Ted

HI, I understand your concern, this has been manufactured for this purpose with a max 150 bar pressure - it compresses on a ploughing machine if it hits an obstacle. These things work and are built for that exact purpose. My point is that I want to understand the workings, its a self contained...

I used bar initially , I assume that you can just multiply answer by 150 to get result for 150 bar? 42729.5 Newtons= (100000 pascals x .005027m2) x (172/172 -170) The first set of brackets I thought are basic SI units e.g pascals, m², & Newtons. Or am I missing something .Is it as simple...

Hi JBA, now that I have worked leverage problem out I am now checking forces along the arc - but this gives me massive numbers. The ram has been using 150 bar so using your formula I use the initial 1 bar pressure 42729.5 Newtons=(100000 x .005027m2) x (172/172 -170) This...

Hi Thanks guys for your responses. It's giving everything a lot more clarity. Yep I probably need to put my equations in excepted format as well to tidy things up. Thanks Brett The circle method has really been a great help (cad makes if super easy)

Hi, This is another machine we have, with all the info I have received thus far regarding leverage. Would I be correct in saying the output force is 177/574 x 5.1 ton giving aprrox 1.5 ton (F2). The red rotates anti clockwise at the pivoit - this I would assume becomes harder because the...

Hi, Thanks for your help, I'm in Australia and working with agricultural equipment and farmers. I'm just using KG's because that is all the layman wants to hear or understand as it will be discussed in those terms as opposed to ;classroom', rough and ready probably. thanks' Ted

This might be easier to read

Hi, Thanks for your replys, please see attached as it is done on cad. Thanks the circle idea has made it very simple please see attached, output F1 I get 740 kg and F2 670 kg. I need to get this too 2500 kg... Thanks That has explained it well if I have it right.

is this the right way to tackle it.

JBA did a stick diagram in another thread that may be easy to understand. thanks

Hi This is a simplified force diagram of a problem I have of some machinery. (shown in same thread) I am trying to get the output force a My calcs gave me 1.5 ton for A and 612 kg for B diagram. when the leverage is not a straight seesaw - do you take arm length from pivot or is one arm...

IN caLc 1 I worked out it would take 2.4 tonne at tip to start to compress cylinder. Calc 2 I worked out 612 kg to start to compress cylinder Calc 3 - it would be 1.5 tonne min to compress past the 25 degree just using distances from fulcrum and class 1 and 2 levers

HI JBA, Thankyou, I did some rough calcs after dragging out my old engineering mechanics book. I will do a sketch to see if my calcs agree, I would be happy to see yours. I think I got 750 kg for the first action hinge point, I'm dubios if was even that as it would not break hard ground at all...

Hi Thanks for your reply. It has been a great help. I was confused with the numbers initially than realized we used 100 and 150 bar. Bring it up to approx 5 ton and 7.5 ton. This now makes sense. I will now move onto the 2nd part of the problem which is the leverage. I might post some more...