Gum Wrapper Survival Hack - Full Explanation
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- čas přidán 26. 09. 2023
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Why do you need to cut a gum wrapper to start a fire with a battery? This question actually came from the CZcams comments. I hope you enjoyed this unnecessary dive into gum wrapper fire-starting circuits. This is a simple but beautiful example of basic circuit knowledge. Let me know if I missed anything. Thanks for watching! - Věda a technologie
Youre wrong about the heat increasing resistance leading to even more heat, that would be positive feedback and cause thermal runaway. It causes negative feedback and less power. For example, a regular old incandescent lightbulb will have an intush current of about 10x that of the normal operating temperature after a coupe seconds. Assuming your voltage source is ideal ( can supply infinite current, or in practice enough for the circuit) the lower the resistance the more power will be disapated.
P=V²/R
This can be confusing as modifying one part of a strictly resistive circuit will effect the other parts. Am easy way to think about it is 0 resistance is a short circuit, infinite resistance is an open circuit. As another comment pointed out, you should also look at the internal resistance of the battery- when included in your circuit it is easy to see that you get max power transfer when resistance is matched (max power transfer theorem).
I think we are actually agreeing (on the negative feedback part), but perhaps I did a poor job of explaining what I meant by the "positive feedback" that I drew (I'll discuss that later in this comment). However, me ignoring the internal resistance of the battery could be a valid concern. I dismissed it for simplicity because I assumed that the gum wrapper resistance (load) was already greater than the internal resistance of the battery (I think around .2 ohms?). If so, then the dissipated power in the gum wrapper (load) would be decreasing with any increased resistance caused by making it skinnier in the middle (in agreement with the MPTT). In contrast, if the gum wrapper has a lower resistance than the internal resistance of the battery, then increasing the resistance of the gum wrapper (load) by cutting it would increase the power to the load (MPTT) until the resistances are equal, and it would make the analysis I did of the simple circuit in the video irrelevant (and I would therefore delete this video because it doesn't make sense to even do this analysis without the internal battery resistance, even though the math checks out). If anyone has the instrumentation to measure an Extra gum wrapper with the folded edges part removed (as shown in the video), let me know what resistance you measure. My multimeter can not accurately measure below 10 ohms. Hopefully someone has that specialized equipment to do so. In hindsight, I should have spent more time validating that assumption before making this video, but hopefully it's fine.
Anyway, assuming the internal resistance of the battery can be ignored for the purpose of this video (and therefore the simple analysis shown still makes sense to do):
I was not trying to say that it is positive feedback for the overall circuit. The total power dissipated does decrease as R increases (P=V^2/R). I used the made up adjective of "concentrated" to just mean the ratio of the power dissipated only by the skinny section divided by the surface area of that skinny section (which is what I label P/A). It's that value that increases even when the overall gum wrapper dissipated power is decreasing. It's a larger percent of the decreasing total power, and is dissipated in a smaller surface area. I stated this in the video, but just wanted to reiterate.
Let me know if I'm still misunderstanding what you mean or if you think I'm misunderstanding any of the concepts. I studied Mechanical Engineering, so I'm obviously not as familiar with some of these topics, which could mean I have delusional confidence that I understand lol. I'm going to pin this for visibility to hopefully get an answer about the load resistance. I really appreciate your thoughtful response! and look forward to hearing from anyone else!
@@JaDroppingScience I agree with mostly everything you said here, although I didn't read it thoroughly. I will read it all the way through and provide any additional thoughts I have, but for now I will clarify why I said you were wrong on one point:
The V²/R does apply to the entire circuit as well and any individual component. We can always draw a box around multiple things and consider them as an individual component con convenience, but that only tells us about them collectively, and each peice still behaves in its own way. Perhaps I misunderstood what you said, I thought you said that as the resistor heats up, Resistance goes up and even more additional heat will dissipate. While I have found that to be true in general, of course that is not always true.
I wasn't thinking about the internal resistance of the battery ( and am not familiar with typical values) . If you model just the source with resistance and the entire load as a resistor, you can plot the voltage / power / current though the load of a carried resistance and see it will look like an upsidedown parabola shape. Sweep the load resistance value up from 0 Ohms and the power dissapated out the load will start at the short circuit output ( see Norton equivalent circuit) , will peak at matched resistance, and then go back down to an open circuit value. So it is indeed possible that raising the resistance can result in more power burned in that component. We have to reevaluate the entire circuit to know for sure, I just assumed it probably wouldn't matter enough to effect the part that burns in a positive way, we won't know for certain unless properly analyzed.
So in summary, I should have said that it is generally the case that higher resistance will burn less power, and this property is negative feedback, which helps keep it from going into thermal runaway. As long the voltage drop across the bridge is relatively the same hot vs cold, then less power will be consumed.
@@JaDroppingScience I may just have to break out the scope and help you, I'll need to procure the gumwrapper.
If you want to make an accurate measurement of low resistance, one simple and easy way you can do so with a Wheatstone Bridge Circuit.
Thank you for your thoughtful response.
@@onradioactivewavesJa was talking about the resistance in the thin part
@@mr.cauliflower3536 ya I know. My point still stands that unless the higher hight => higher resistance caused the circuit to be impedance matched to the "source" ( and I doubt that, but it is possible) or at least moved in that direction enough to overcome the increased resistance, then the heat/energy dissipated will be less.
So I was saying that that statement he made is generally wrong but not always. I may have misunderstood or perhaps he mispoke, I think we are in agreement about the issue nonetheless, so no sense in splitting hairs over something that was not intended to be said.
dang, I've been trying to get a battery tree spawn for *hours!*
There actually not all that rare you just need to play the game long enough for them to grow (12 in game days) I've made very similar mistakes.
@@greenlight5307 Even if you check the trees every day for 12 days there's a chance you won't find any batteries. For me, someone else always gets there ahead of me and takes them all. There have been times where I've had to stay up all night waiting for the batteries to spawn so I can grab them before someone else can get there first, and even then there's been times where I haven't gotten them because someone showed up at the last minute and shoved me away.
Try spawning in Thneedville
@@greenlight5307bro I’ve been waiting 69 days
Honestly tho
"You've probably seen this survival trick before"
Nope, the first few seconds of the video seemed like alchemy before the explanation lol
“but if you want to see me go into more detail”
*sees a bunch of math*
“fuck no”
0:09 i was expecting you to say the classic one line "do not try at home"
so it would be safe to try it in school or somewhere else 😂
How much battery and/or gum wrapper do I need to start a fire that can burn a brick and wood structure?
Man's out here asking the right questions
you dont need anymore, what you need is something easily flammable like newspaper to hold the fire until the wood starts catching on fire. I recommend trying to set the curtain alight first.
you got yourself an additional sub.
your vids are ez to understand and straight to the point.
thnx mate.
I see you have less views on these than the other videos. If you enjoy teaching cool stuff like this to people, you absolutely should continue. 20 thousand curious folks, after all.
Nice concise well made video
Love your stuff!
Yooooo
Nice check. Works like a charm.
While the current decreases, the current density in the skinny region greatly increases. A high current density is what leads to a material overheating and subsequently igniting.
Think of it this way: a very high current of 10 Amperes could easily be transmitted via a copper wire if the wire was fat enough, thus limiting the current that each unit of cross-sectional wire transmits. But, if the wire is skinnier, then each unit of cross-sectional wire will be overwhelmed and overheat.
Good job bro 👌👌
Item spawn 🤣 Brilliant! Congrats on 1 million! Is discord still a thing? I thought I was on your server but I can't find it.
Thanks! and I closed it down because I didn't do a great job keeping it active. If I can get to a place where I can make it how I originally intended, I'll re-launch one. Sorry about that.
@@JaDroppingScience I run several discord servers. I'm happy to help
Bro burned his house for da intro
Greater video, you didn't touch on matching the batteries internal resistance maybe
Excellent point, I assumed the load resistance (gum wrapper) was already greater than or equal to the batteries internal resistance. I should have verified that assumption. If anyone has a specialized multimeter or way to measure small resistances, if you could let me know what the gum wrapper resistance is (with the folded parts cut off as shown in the video), that would be great. If it's less than the battery internal resistance (between .1 and .9 ohms), then I'm going to delete this video.
For more details why: see my response to the pinned comment.
@@JaDroppingScience thanks for the answer! unfortunately I don't have a mohm meter, but maybe someone from the audience does :)
My hypothesis before watching is no (ill add edit after)
I was wrong 😔
Ohm's law applies to static systems. Since the circuit is changing with time, you need to bring electrodynamics into it. Saying "the heat increases, therefore the resistance increase, therefore the power density increases, therefore the heat increases" is circular logic. You used the fact that the heat is increased to deduce the heat is should increase. I'm not an expert in this topic, I'm just throwing in my two cents.
LOL, inches. And you call yourself a scientist! xD
You should see my bottle opener video, I used Newton-Inches for some calcs lol.
You missed talking about the impedance of the battery in your equations.
What's missing is an explanation of why we're doing this? Is it an improvised fire starter?
I believe it is meant to be either a fun party trick or an improvised fire starter. Both are possible.
Yeah technically a survival hack/trick, although I just think it's cool to try to compare simple theory to the reality to see if it explains what's happening beautifully or in ways that are intuitive.
EARLY
Seventh!
First
No, second
@@Blu5-_-nah he first comment
@@erner_wisal nope