Equal Percentage characteristic can be identified by equal increments of X axis that yield equal percentage increments of the y-axis.

Whereas, in a linear characteristic, equal increments of x-axis yield equal increments in the y-axis.

Linear curve is simple as equation is just y=x.

I need help for the equation for equal percentage curve.
Either equation or how to generate it in excel will be gr8.

I have posted this picture of both linear(1) and equal percentage curve(2)  which will tell you what I am asking.
Thanks in advance

The equation is y = exp(x) i.e. exponential of x. Try it it will give you equal percentage increment of y for equal increment of x.

I dont know the exact equation. If you are using this on flow control charecteristic you can use exponential type of equation but not exactly Y=e^X. You can use y=a+b*(e^(x^n)-1) where you can find the values of 'a' and 'b' depending upon your boundary condition/initial conditions i.e based on the type of parameters used too. 'n' gives the smoothness of curve for example square or cubic.

I am not the mathematician but I like to apply mathematics.

Parbatya

it's very much akin to y=mx+c (where m =slope and c=intercept), except that the slope is exponential

y=e^(mx)+c 

where, m=percentage increase
c=intercept (at the y-axis in your case)

Parbatya, its for flow characteristics. 


with linear, y= x. (slope =1)
As I am comparing with slope 1, with my equal percent curve, will it be n (or m) =1

I am getting the exponential equation you guys suggested me. But I am not just able to get that curve. For e.g. when x is 40%, i am not getting y=10%. see the graph in the first post. Suggest me the pattern.



Please help

Any equation that satisfy the relation f(x+a)= (1+a/100) f(x), where a is any fixed real number, has that property.

Any function f(x) that satisfy the equation f(x+a)= (1+a/100) f(x), where a is any fixed real number, has that property.

when x is 40%, i am not getting y=10%

That's not equal percentage then. for 40% increase in x, y should also increase by 40% for an equal percentage curve (as per my understanding).

I am guessing you are looking for modified parabolic or hyperbolic, where small increase in y leads to greater increase in x.

y=K.exp(mx)+c

I don't have access to Excel or Matlab right now to verify. Try to apply yourself. Try playing around with K to get the percentage change. I am guessing, it should be 1/4 in your case assuming m to be 40.

well, it is exponential curve. I now realise that I was trying to be specific to that plot with generalized data. Parbatya ( I know you), I will contact you later on to be more specific and clear my confusion. I know you can help me.

lootekukur, well, it is when opening is 40%, flow is 10% (in the plot in first post). The book I am going through says it is equal percent(not modified parabolic or hyperbolic), but I cannot deny that possibility too. I will try to use that. Thanks

That's not equal percentage then. for 40% increase in x, y should also increase by 40% for an equal percentage curve (as per my understanding).

Yeso bhayo bhane ta linear bhai go ni haina ra?

For linear function , y = x since the slope is 1 ,

Now, for exponential can't you do,

 y = Ar^x  

where A=10^b and r =10^m  (m and b are the slope and the intecept of the line). I tried to plot on the matlab before post anything here but I couldn't since another program is running in my machine.

The formula is y/100=K^(x/100-1). K can be any constant which will change the shape/rate of the curve. Here is a check with K=20.

Here is example for 10% increase.
x                 y            percentage increase
0                 5.00             25.89
10               6.75             25.89
20               9.10             25.89
30              12.28             25.89
40              16.57             25.89
50              22.36             25.89
60              30.17             25.89
70              40.71             25.89
80              54.93             25.89
90              74.11             25.89
100            100.00           25.89

Here is example for 15% increase.
x                 y            percentage increase
0                  5.0                   36.2
15                7.83                  36.2
30               12.28                  36.2
45               19.25                  36.2
60               30.17                  36.2
75               47.29                  36.2
90               74.11                  36.2

Use this equation for your particular problem of 'CONTROL VALVE CHARECTERSTIC'

Y = 0.02 * EXP(3.91X) 

EXP means Exponential

X and Y are in fraction, i.e. 0.1 means 10 %

And the result looks like

0.00	0.02
0.10	0.03
0.20	0.04
0.30	0.06
0.40	0.10
0.50	0.14
0.60	0.21
0.70	0.31
0.80	0.46
0.90	0.68
1.00	1.00

Hope it helps

 

Parbatya

Here is the graph for K=20. You can draw the graph in excel using the formula.

dyamn...

_x

For 40% opening, 10 % flow, K should be 46.42

v2001,

is y=10% when x=40% all that is needed? that might be sufficient to fit the curve (mathematically) but doesn't sound technically right to me. there must be other requirements. if not, it's a strange problem.

it took me a while to figure that when you said, x is 40%, that's a value not percentage change.
an equal percentage curve is such that for the increase of x by same amount (in number, not in %), y changes by equal percentage. that is nothing but an exponential curve.

y=exp(x)
 
for your case, the generic curve y=Kexp(nx) +C should work. it's just the matter of finding the right values for constants for your problem IMO.


khaobadi,
it's not 10% or 15% increase, but 10% or 15% open (valve that is, in this case). And your second example does not yield same percentage change in y (flow). That is not equal percentage.

Parbatya's example also doesn't entail equal percentage curve IMO (by definition) although he was able to yield y=10% for x=40%.

Wait till the evening when i have time, i will get this done if you are not able to solve it till then.

LooTe, as far as my understanding of the equal percentage valve it is the equal percentage increment of flow for a equal lift.

From http://www.boacalgary.com/artvalve.html

Equal Percentage Characteristic - In a valve having an equal percentage characteristic, like movements of the valve stem at any point of the flow range changes the existing flow an equal percentage regardless of the existing flow.

Example:

Suppose a valve stem has been lifted 30 percent of its total lift and the flow at this time is 3.9 gal/min. Now assume that the valve opens an additional 10 percent of its full travel and that the flow increases to 6.2 gal/ nun or a 60 percent increase. Next, suppose that the valve stem moves an additional 10 percent so that it is now 50 percent open. The flow now will be 10 gal/min or another 60 percent increase in flow.

CT,  prolly in some civil/mechanical engineering problem. i am guessing--- some dam/turbine problem. but i don't know for sure. i am just interested in the mathematical bit.

------
yes khaobadi. that is exactly what i am saying, equal percentage flow for equal increment in valve opening.

but that example gives '59.something' percentage change in the first case and '61. something' percentage change in the second. i am sure, there is ought to be some tolerance, but that's just a tad bit big IMO. I am not sure if its tolerable for this particular problem (mechanical/civil)

and your second example of your first post gives '36.6' percentage' initially and '56.something' later on.

parbatya's solution also gives range of percentage change -- from 33% to 60%

i am thinking, there are other requirements to this problem (for it to be complete so as to get a unique answer). else, there could be myriad solutions and values for the constants --none of them giving a perfect equal percentage solution.

or it could just be that he wanted to draw just a close-to-accurate graph and that's it.

let's hear more from v2001.