Recent content by Geordielad

Erobz, I have figured out how to write the ##f (x) ##you mention in post 29 using the law of cosines and since I know all 3 sides. ##f(x)=\cos\left(\frac {x^2+a^2-b^2}{2xa}\right)## This gives me the angle designated ##sin beta## in post 29. for all the intermediate positions of the cylinder...

As I understand it the terms are interchangeable.

Thanks.

Thanks, I do see the formula and I understand angle##\theta## but I do not understand what angle##psi## is. I think I must have ##\cos\psi## to figure out what "h" is and I need "h" to figure out what ##F_s## is, which is my ultimate goal. I have inserted my figures to ##\cos\psi## and "h" it is...

Hi I have a very similar problem to this just different geometry and I am trying to follow your example above so that I can determine the ##F_s## force for my arrangement. Could you please explain what ##\cos\psi## refers to as it is not shown in the diagram. Thanks

Is there an easier way for me to calculate Fs?

It is to help a friend who has even less math knowledge than me. I am 74 years young and my math is rusty to say the least. My apologies but I do not understand function of (x). and The ideal gas law is way over my head. Could I ask you to recap how to find Fs we got so far and then I lost the plot.

What I was trying to say was that when the cylinder is compressed there is less volume therefore a higher pressure than when extended. Your suggestion to solve Fs analytically sound good but I don't understand ##\sin\beta=function (x),\cos\theta=g (x), F_s=q (x)## could you please explain how to...

So if I understand what you are saying I should be using ##F=P*A## where F is cylinder force, P is pressure in cylinder, A is cross sectional area of piston, ##F_1## is extending force,##F_2## is compressing force, ##V_1## is cylinder volume extended, and ##V_2## is cylinder volume compressed...

$$\frac {2c\sin\beta} {e\cos\theta}$$

I will rewrite using \frac for the ##2c \sin\beta##section. The gas spring manufacturer gives the formula ##F*E = S*J ## where F is force, E is the force moment arm, S is door weight, J is weight moment arm. Then ##F=\frac {S*J} {E}## with this I used the spreadsheet to calculate F for...

I have create a spreadsheet to compute all the trig functions to establish the fixed and moving points so that I can make iterations for the moveable mounting bracket (also different gas spring lengths). When I insert the FH formula below I get a reasonable figure to close an open door but...

Thanks Erobz

Okay solving for F_H $$0.5 (e cos \theta) W + (e cos \theta) F_H - 2(c sin\beta) F_s=0 $$ $$0.5 (e cos \theta) W + (e cos \theta) F_H = 2(c sin\beta) F_s $$ $$ (e cos \theta) [ 0.5 W + F_H ] = 2(c sin \ beta) F_S $$ $$ [ 0.5 W + F_H ] = 2(c sin \ beta) F_S / (e cos \theta) $$ $$ F_H = (2(c sin \...

$$ e F_H cos \theta + \frac 1 2 e W cos \theta -2 c F_s sin \beta =0 $$ is this correct ?