square root of the sum of the squares...

[tris-]

im pretty competent so far in calculating relative uncertainty and the probable relative uncertainty to work out the error in equations such as

a = b+c

a = (bc)/ d

but im stuck when it comes to doing it for an equation like

a = (b-c)/d

what im unsure of is do you use the same method of error propogation to calculate the relative uncertainty of a or is there some other method as the equation involves both subtraction and division rather than the usual multiplication and division.

 

[swords]

Normally this is where i leap in and save the day...however I am not a math person
Even standard deviation gives me the creeps

 

[tris-]

Normally this is where i leap in and save the day...however I am not a math person
Even standard deviation gives me the creeps

S.D is the measure of the width 88.2% up the height of the graph

 

[Keitan]

im pretty competent so far in calculating relative uncertainty and the probable relative uncertainty to work out the error in equations such as

a = b+c

a = (bc)/ d

but im stuck when it comes to doing it for an equation like

a = (b-c)/d

what im unsure of is do you use the same method of error propogation to calculate the relative uncertainty of a or is there some other method as the equation involves both subtraction and division rather than the usual multiplication and division.

a = (b-c)/d

i dont know what u mean and i dont care either but i know to shorted that down is a = b-c divide by d

 

[Neffneff]

a = (b-c)/d

i dont know what u mean and i dont care either but i know to shorted that down is a = b-c divide by d

lol?

 

[Overdriven]

[Removed] =( I got lost.

 

[tris-]

a = (b-c)/d

i dont know what u mean and i dont care either but i know to shorted that down is a = b-c divide by d

wow erm, thanks mate.

but for future reference youve made the whole thing longer by adding in "divided by"

a=(b-c)/d

is less characters

[Removed] =( I got lost.

THINK MOTHER FUCKER!

 

[crispy]

im pretty competent so far in calculating relative uncertainty and the probable relative uncertainty to work out the error in equations such as

a = b+c

a = (bc)/ d

but im stuck when it comes to doing it for an equation like

a = (b-c)/d

what im unsure of is do you use the same method of error propogation to calculate the relative uncertainty of a or is there some other method as the equation involves both subtraction and division rather than the usual multiplication and division.

well if your calculated value a is derived from the 3 values b, c and d each with their own uncertancy you'd calculate a's uncertancy like this:

in your particular case (big letters are your variables, small uncertancy value):

sqrt( (((d/dC (B-C)/D))·c)^2 + (((d/dB (B-C)/D))·b)^2 + (((d/dD (B-C)/D))·d)^2 )

Well at least thats how ive been calculating uncertancy for my results last time (was the first time also so D)

edit:
Hmm read it again... its prolly not the sqrt thing youre not sure about but how to differentiate the (B-C)/D) part right?

Like this:

d (B-C)/D) = 1/D
dB

Shitty formatting... :>

 

[Alan]

a = (b-c)/d

i dont know what u mean and i dont care either but i know to shorted that down is a = b-c divide by d

and thats actually wrong, you do multiplication and division before addition and subtraction unless there are brackets

a = b-c/d (is not the same as) a = (b-c)/d

Anyway, I also have no idea what the hell is going on here I got a GCSE grade C for maths.

 

[crispy]

f(x,y,z) = your function (any number of variables)
fy' = 1st derivative function of f according to y
fx' = 1st derivative function of f according to x
fz' = 1st derivative function of f according to z
az = your uncertanty for z etc...

sqrt( (fy'·ay)^2 + (fz'·az)^2 + (fx'·ax)^2)

ofc, if you use more a function depending on even more different uncertanties you'd have to expand this one accordingly

damn im shitty at explaining

 

[[NO]Subedai]

thats quite low level algebra, and i could never handle algebra, i allways found it way to hard to figure out, when i had my gcse exam i got an A* primarily through figuring the algabra out by doing process of elimination.

example

A is 1
b is 2
c is 1
d is 1

 

[tris-]

lol no no no. i think youve misunderstood me. i know algebra fine well.

i know that for an equation that contains + and - or divide and multiplication, i know that the probable relative uncertainty for the answer to the equation is calculated by the square root of the sum of the squares.

now when an equation involves, for example, - and divide and addition, how is the probable uncertainty calculated for the answer to the equation?

do you still use the square root of the sum of the squares to calculate the probable uncertainty, or is another method used?

to clear -

i dont need help re-arranging algebra

i dont need help subsituting letters for numbers

i know what BODMAS is

Variable	Absolute Error	Relative uncertainty	Power in formula	Effective r%	Square of relative	
a	13.61	   0.02	                   0.147	                   1	                    0.147	0.022	
b	114.5	    1.6	                   1.397	                    1	                     1.397	1.953	
c	19	     0.8	             4.211	                      1	                       4.211	17.729	
							
					                                                                Sum	19.70	
				                      Probable relative uncertainty		4.44	%
y	82.02	± 4.44%

so here you can see what im doing, and that i know how to work out the uncertainty for the equation y = ab/c

formatting fucked but just read along and you should be able to fit the numbers in the right column

 

[Alan]

to clear -

i dont help re-arranging algebra

i dont need help subsituting letters for numbers

i know what BODMAS is

is it..... 42 ?

 

[crispy]

Well, i can show you how i do it using your examples. Mind you, only recently ive had to do uncertancy calculations :>

First case:

F = A + B ,B got uncertancy b and A got uncertancy a.

Then id calculate the relative uncertancy for F like this:

f = sqrt( (a)^2 + (b)^2 )



Second case:

F = A^2 + B^2

would look like this:

f = sqrt( (2·A·a)^2 + (2·B·b)^2 )



Third case:

F = A^2·B + (B^2)/C + 1/C

would give this uncertancy:

f = sqrt( (2·A·B·a)^2 + (((2·B)/C + A^2)·b)^2 + ( (-(B^2 + 1)/C^2)·c )^2 )

---

Hmm not sure im even onto your problem

 

[tris-]

rofl. right, the question im asking is how do you find the uncertainty of y in the following equation -

y = (b-c)/d

do you use the square root of the sum of the squares method as shown in the above example.

OR, OR

is there some other method to calculate the uncertainty of y when both division AND subtraction is used?

as opposed to

subtraction AND addition

multiplication AND division

as all i know is when the equation involves a mix, i dont know the correct method to calculate uncerainty.

and i really cannot explain it anymore than that

 

[Reformed]

Blimey! Reaches for a bottle of whisky. Nasty stuff is maths by the looks of it. You sure it is good for you to think in them terms? Keeps you sharp I bet. Probably a bit too sharp. Can you sleep at nights?

 

[tris-]

Blimey! Reaches for a bottle of whisky. Nasty stuff is maths by the looks of it. You sure it is good for you to think in them terms? Keeps you sharp I bet. Probably a bit too sharp. Can you sleep at nights?

if its bed time and i still have equations i have to finish them all or i cant sleep

day 3 and counting now

 

[Sharaft]

thats a nice easyone, you solve (b-c) then divide it on d... then you should have so you got y=bc/d... if d is an unknown you got the answer right there..

what about x(x+3)^2=29

get this one tris?

 

[MaCaBr3]

This is where I jump in this thread, make a useless post and get out again.

 

[Mojo]

Number are for adding, letters are for writing. Stick to that and you will be fine, try to fuck with it and well............

 

[Sharaft]

Number are for adding, letters are for writing. Stick to that and you will be fine, try to fuck with it and well............

but when you put numbers and letters together you get a VERY good combination that can blow your brain out, and still survive!

 

[Ezteq]

i dont have a clue what your talking about but..

wheres my goddamned newspaper clipping tris!!!???

I BEEN WAITING BY ME LETTERBOX FOR A WEEK NOW

A WEEK MAN!!!!

im starting to get cramp in my left buttock

 

[Mojo]

but when you put numbers and letters together you get a VERY good combination that can blow your brain out, and still survive!

U r0ck5

 

[Sharaft]

U r0ck5

totally

 

[tris-]

thats a nice easyone, you solve (b-c) then divide it on d... then you should have so you got y=bc/d... if d is an unknown you got the answer right there..

im not asking how do you solve an equation, im asking how to propogate errors and calculate the probable uncertainty for an equation that involves subtraction and division.

no offence but unless you know what the square root of the sum of the squares method for error propogration is, then you probably dont know what im asking for help with

ez, everytime i think i got a chance, i find i have shit loads of work to do :/

 

[crispy]

rofl. right, the question im asking is how do you find the uncertainty of y in the following equation -

y = (b-c)/d

do you use the square root of the sum of the squares method as shown in the above example.

OR, OR

is there some other method to calculate the uncertainty of y when both division AND subtraction is used?

as opposed to

subtraction AND addition

multiplication AND division

as all i know is when the equation involves a mix, i dont know the correct method to calculate uncerainty.

and i really cannot explain it anymore than that

Well if ive understood this correct you try to calculate a value y from 3 other values? And then you want to know the uncertainty of y since the 3 values you calculate y from also got an uncertainty?

If so you'd still use the square root of the sum of the squares method basicly. Except with a little modification. The squares you want to sum and take the square root of is calculated a little differently.

Normally when you just got Y = A - B youd get the y uncertainty like this:

sqrt( a^2 + b^2 ) --- small letters denote uncertainty for A and B

Now when you got your function Y = (A-B)/C you will have to square the derivative of Y according to each of your A,B and C values and multiply it by your uncertainty. Like this:

sqrt( ((1/C)·a)^2 + ((-1/C)·b)^2 + (( -(A-B)/C^2) ·c)^2 )

It is actually the same you do normally because the function Y = A + B looks like this when you derive it according to A and B:

Ya' = 1 --- Ya' is Y derived according to A
Yb' = 1 --- Ya' is Y derived according to B

Put into the square root of squared method it looks like youve seen it normally:

y = sqrt ( (Ya'·a)^2 + (Yb'·b)^2 )
also seen as:
y = sqrt ( (1·a)^2 + (1·b)^2 )

-----

The function you wanted to calculate the uncertainty for is derived like this:

Ya' = 1/C
Yb' = -1/C
Yc' = -(A-B)/C^2

When we put this into the squareroot it looks like this:

y = sqrt ( (Ya'·a)^2 + (Yb'·b)^2 + (Yc'·c)^2 )

Then just insert the Ya' etc values instead to get the function.

So to conclude, you would still use the square root of squares method as youre used to, but you'd square some different values. The derived values times their uncertainty.

 

[tris-]

cheers

 

[Haggus]

dude go ask your teacher that will solve the problem

Tho my teacher never new the answer to my maths ;/

 

[tris-]

dude go ask your teacher that will solve the problem

Tho my teacher never new the answer to my maths ;/

believe it or not its usually quicker and a lot easier to ask people here

 

[rayzor]

a = (b-c)/d

i dont know what u mean and i dont care either but i know to shorted that down is a = b-c divide by d

lol
Well, that brightened up a dull Friday!