# Solving Filament Failure Mode with Maths

• CWatters
In summary, when it comes to the filament in a lamp, the book explains that the thinner part of the filament is hotter due to evaporation occurring faster there. This is because of the smaller cross-sectional area and higher resistance. However, when attempting to show this mathematically, the power in the thinner part decreases with an increase in resistance, contradicting the book's explanation. However, this discrepancy may be due to the physics of the filament, where the decrease in surface area also makes it harder to dissipate power. Additionally, it is important to note that when R2 is small compared to the total resistance, the current through the combination will be roughly constant, and the dissipation in R2 will be proportional to its resistance.
CWatters
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## Homework Statement

Not homework but this seems the best place to post this. In a private conversation someone posted this explanation from a book on how a lamp filament fails..

when it comes to the filament in a lamp the book said the thinner part of the filament is hotter than other parts and evaporation occurs faster there because of smaller cross sectional area and higher resistance. This makes the thinner part even thinner until it breaks.

I'm having some trouble doing the maths to show that the thinner higher resistance part gets hotter...

Ohms law

## The Attempt at a Solution

[/B]
Imagine the filament is made up of two resistors in series R1 and R2 (See diagram). Let R2 be the part that's getting thinner for some reason...

The voltage on R2 is given by the potential divider rule..

VR2 = V * R2/(R1+R2)

The power dissipated in R2 is

WR2 = (V * R2/(R1+R2)2/R2

If you expand and simplify you get..

WR2 = (V2 * R2) / (R12 + 2R1R2 + R22)

So if R2 increases the power in R2 decreases due to the R22. That should reduce the temperature in R2 which is the opposite of what the book suggests. Have I made an error?

I suspect the problem is to do with the physics of the filament. eg the thinner part has less surface area so although the power dissipated in that bit is lower the surface area is also lower so it's harder to dissipate the power.

CWatters said:

## Homework Statement

Not homework but this seems the best place to post this. In a private conversation someone posted this explanation from a book on how a lamp filament fails..
I'm having some trouble doing the maths to show that the thinner higher resistance part gets hotter...

Ohms law

## The Attempt at a Solution

[/B]
Imagine the filament is made up of two resistors in series R1 and R2 (See diagram). Let R2 be the part that's getting thinner for some reason...

View attachment 85636

The voltage on R2 is given by the potential divider rule..

VR2 = V * R2/(R1+R2)

The power dissipated in R2 is

WR2 = (V * R2/(R1+R2)2/R2

If you expand and simplify you get..

WR2 = (V2 * R2) / (R12 + 2R1R2 + R22)

So if R2 increases the power in R2 decreases due to the R22. That should reduce the temperature in R2 which is the opposite of what the book suggests. Have I made an error?
If R2 is small compared with the total resistance, the current through the combination will be roughly constant. The dissipation in in R2 will be I^2 x R2, so it is proportional to the resistance of R2. In your formula, R2 appears in both top and bottom, so your statement is not quite accurate.

Ah yes I'm forgetting that the square of a number less than 1 is smaller than the original number.

If R2 is the part getting thinner then, assuming R2 is same length as the rest of the filament R1, R2 will be bigger than R1, and
power in R2 = i*R2 = [V/(R1 + R2)]R2 which is > [V/(R1 + R2)]R1.

Question for you: why is R2 bigger than R1?

## 1. What is filament failure mode?

Filament failure mode is a term used to describe the various issues that can arise when using 3D printing filament, such as clogging, nozzle jams, and inconsistent extrusion.

## 2. How can maths help solve filament failure mode?

By using mathematical calculations and algorithms, we can analyze the printing process and identify patterns and factors that contribute to filament failure mode. This allows us to make adjustments and improvements to prevent these issues from occurring.

## 3. What specific mathematical concepts are used in solving filament failure mode?

Some of the mathematical concepts used include flow velocity, temperature, material properties, and pressure. These factors are all interrelated and can affect the printing process and result in filament failure mode if not properly understood and controlled.

## 4. Can using maths completely eliminate filament failure mode?

While using maths can significantly reduce the occurrence of filament failure mode, it is not a foolproof solution. There are still external factors such as environmental conditions and machine maintenance that can affect the printing process and lead to filament failure mode.

## 5. Are there any mathematical tools or resources available to help with solving filament failure mode?

Yes, there are various mathematical models and simulations specifically designed for 3D printing that can assist in analyzing and predicting filament failure mode. Additionally, there are also software programs and online calculators that can help with calculations and optimizing printing settings.

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