How do you calculate the uncertainty in T using graphs?

[heavystray]

i have an equation like this:

Given m(B-V)= 1.2 +- 0.2
How do you calculate the uncertainty in T? (btw, I solve T using graphs by finding the intersection point)

My idea was first to calculate T when m(B-V) =0.2, and I then calculate T when m(B-V)= 1.2 + 0.2(its uncertainty). and then find the difference between the two T. Your help would be greatly appreciated

 

[kuruman]

Your attachment cannot be viewed.

 

[heavystray]

Your attachment cannot be viewed.

ughh thanks for pointing that out, sorry!

 

[jbriggs444]

There is the simple way -- assume that the solution for T is monotone in m(B-V). Evaluate T for m(B-V) = 1.4 and for m(B-V) = 1.0. See what the range is.

Evaluating the error range by looking at a partial derivative of T with respect to m(B-V) and then multiplying by the error bound on m(B-V) sounds like way too much work and may not even be accurate.

 

[Khashishi]

Take the derivative of both sides. You get something of the form:
dm(B-V) = (...) dT
where I'm too lazy to figure out the part in parentheses.
Then the error is
dT = dm(B-V) / (...)
then just let dm = 0.2

This assumes that the error is small, so that you can treat the error in T as linearly proportional to the error in m. This will fail if the error is actually not small.

 

[kuruman]

You can invert the equation and solve for T in terms of m(B-V). Then do standard error propagation. Use
$$\frac{{e^{\frac{a}{T}} } } {{e^{\frac{b}{T}}}}=e^{\frac{1}{T}(a-b)}$$

 

[heavystray]

There is the simple way -- assume that the solution for T is monotone in m(B-V). Evaluate T for m(B-V) = 1.4 and for m(B-V) = 1.0. See what the range is.

Evaluating the error range by looking at a partial derivative of T with respect to m(B-V) and then multiplying by the error bound on m(B-V) sounds like way too much work and may not even be accurate.

what do you mean the for T is monotone? so the range would be divided by two right? thanks for the reply

 

[heavystray]

You can invert the equation and solve for T in terms of m(B-V). Then do standard error propagation. Use
$$\frac{{e^{\frac{a}{T}} } } {{e^{\frac{b}{T}}}}=e^{\frac{1}{T}(a-b)}$$

Wait, I'm really sorry, the equation supposed to have minus one after e, that's why i can't solve for T in terms of m(B-V) numerically. sorry again for uploading the wrong eq. thanks for the reply

 

[kuruman]

Wait, I'm really sorry, the equation supposed to have minus one after e, that's why i can't solve for T in terms of m(B-V) numerically.

That makes a big difference. Thanks for the clarification.

 

[jbriggs444]

what do you mean the for T is monotone? so the range would be divided by two right? thanks for the reply

A function is monotone if it it always increases when its argument increases. Or if it always decreases when its argument increases. If a function is monotone then it takes on its extreme values at the extreme values of its input. This is a less strict condition than "linearly proportional".

The range would not be divided by two. The uncertainty could be different on each side of the measured/computed value.

WIth an uncertainty that is nearly as much as 20% of the measured value it might be wise to discard the assumption of linear proportionality.

 

[kuruman]

What is the value of T that gives m(B-V) = 1.2? I assume it is an experimental number.

 

[heavystray]

A function is monotone if it it always increases when its argument increases. Or if it always decreases when its argument increases. If a function is monotone then it takes on its extreme values at the extreme values of its input. This is a less strict condition than "linearly proportional".

The range would not be divided by two. The uncertainty could be different on each side of the measured/computed value.

WIth an uncertainty that is nearly as much as 20% of the measured value it might be wise to discard the assumption of linear proportionality.

if we use the above eq. when
m(B-V)= 1.2, T= 3930
when m(B-V)= 1, T= 4420
when m(B-V)= 1.4, T= 3530

so the uncertainty of T= is 4420-3530?

OR
we have to do it separately on both sides?
upper uncertainty= 3920-3530
lower uncertainty= 4420-3920

Thanks

 

[heavystray]

What is the value of T that gives m(B-V) = 1.2? I assume it is an experimental number.

if we use the above eq. when m(B-V)= 1.2, T= 3930
when m(B-V)= 1, T= 4420
when m(B-V)= 1.4, T= 3530

 

[kuruman]

if we use the above eq. when m(B-V)= 1.2, T= 3930

For m(B-V) = 1.2 I got T = 4960 for log base 2.51. The natural log gave me 3714 which is closer to your value of 3930.

 

[heavystray]

For m(B-V) = 1.2 I got T = 4960 for log base 2.51. The natural log gave me 3714 which is closer to your value of 3930.

wait, how do you find the 4960?

 

[heavystray]

A function is monotone if it it always increases when its argument increases. Or if it always decreases when its argument increases. If a function is monotone then it takes on its extreme values at the extreme values of its input. This is a less strict condition than "linearly proportional".

The range would not be divided by two. The uncertainty could be different on each side of the measured/computed value.

WIth an uncertainty that is nearly as much as 20% of the measured value it might be wise to discard the assumption of linear proportionality.

if we use the above eq. when
m(B-V)= 1.2, T= 3930
when m(B-V)= 1, T= 4420
when m(B-V)= 1.4, T= 3530

so the uncertainty of T= is 4420-3530?

OR
we have to do it separately on both sides?
upper uncertainty= 3920-3530
lower uncertainty= 4420-3920

Thanks

 

[jbriggs444]

we have to do it separately on both sides?
upper uncertainty= 3920-3530
lower uncertainty= 4420-3920

Thanks

Assuming that the numbers are accurate, that's 3920 minus 390 to plus 500. Definitely asymmetric above and below. See the link below for some advice on reporting asymmetric uncertainties.

https://stats.stackexchange.com/que...referred-way-to-give-asymmetric-uncertainties