I/P convertor equation finding 2 unknowns

[tommoturbo]

Homework Statement

Basically its all on the attached sheet as i don't know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.








But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA



We know the values of Kc and A and hence can find x and y. Sorry but I can't fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?


(N.B. be sure to have all measurements in S.I units.)

Homework Equations

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x


A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

The Attempt at a Solution

i can't attach any files I am afraid as it won't work?


any help greatly appreciated

 

[berkeman]

Homework Statement

Basically its all on the attached sheet as i don't know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.








But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA



We know the values of Kc and A and hence can find x and y. Sorry but I can't fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?


(N.B. be sure to have all measurements in S.I units.)

Homework Equations

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x


A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

The Attempt at a Solution

i can't attach any files I am afraid as it won't work?


any help greatly appreciated

Attaching files should work. Is it real big or something? Also, it's better to post as a PDF -- less chance of transfering worms etc. You can get a free PDF converter from PrimoPDF if you don't have one already.

 

[Mark44]

Since x + y = 100, you can replace y with 100 - x so that your equation for Po involves only one variable.

You haven't provided enough information for me to figure out what you're doing. E.g., what units is Kc in? I don't understand N/A. A usually represents amperes, but that doesn't seem to be what you're doing.

 

[tommoturbo]

Hi guys thanks for the replies sorry its so vague i did it all on a word doc importing graphs and the equation etc thinking i could post it but the uploader won't work.

the units for the constant Kc are in Newtons but it has an A to the minus 1 after it.


ill try and post it from home it might not work on the pc here.

thanks again for replying.

 

[tommoturbo]

HI guys here is the notes i have which should clear things up i hope

 

[Mark44]

But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x

This formula is different from the one in the attachments. This is the one in your attached files:
P_o = \left( \frac{K_c y}{A x}\right)I_{in} + \frac{F_s z}{Ax}

The way you wrote it above indicates that only A is in the denominators, not Ax.

This problem is a lot more confusing that it should be. For one thing, A is being used for three different purposes, making it difficult to figure out what A represents in a given situtation. One use for A is identifying one of the two straight lines (A and B). Another use is area. The third use is for amperes.

The equation above gives P_o as a linear function of I_in. Since line "A" (better, line 1) goes through the origin, (F_s z)/(Ax) must be zero.

You are given that m = (1/20) mA, which means that the expression that multiplies I_in must be equal to 1/20 mA, so solve the equation (K_c y)/(Ax) = 1/20.

A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

The notation N A^(-1) is the same as N/A, or Newtons per ampere.