1 |
DENIED
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ROLE_USER
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Show voter details
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2 |
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moderate
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Proxies\__CG__\App\Entity\Entry {#1583
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Show voter details
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3 |
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edit
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Proxies\__CG__\App\Entity\Entry {#1583
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Show voter details
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4 |
DENIED
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moderate
|
Proxies\__CG__\App\Entity\Entry {#1583
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|
Show voter details
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5 |
DENIED
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ROLE_USER
|
null |
|
Show voter details
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6 |
DENIED
|
moderate
|
App\Entity\EntryComment {#1612
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|
Show voter details
|
7 |
DENIED
|
edit
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App\Entity\EntryComment {#1612
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…2
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|
Show voter details
|
8 |
DENIED
|
moderate
|
App\Entity\EntryComment {#1612
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|
Show voter details
|
9 |
DENIED
|
ROLE_USER
|
null |
|
Show voter details
|
10 |
DENIED
|
moderate
|
Proxies\__CG__\App\Entity\Entry {#1567
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+body: """
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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}
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date: 2023-11-23 16:39:47.0 +01:00
}
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0 => App\Entity\EntryComment {#1548
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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…2
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Show voter details
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11 |
DENIED
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edit
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Proxies\__CG__\App\Entity\Entry {#1567
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1 => App\Entity\EntryComment {#1683
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+body: """
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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Show voter details
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12 |
DENIED
|
moderate
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Proxies\__CG__\App\Entity\Entry {#1567
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+body: """
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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App\Entity\EntryComment {#1683
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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DENIED
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App\Entity\EntryComment {#1683
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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Show voter details
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16 |
DENIED
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App\Entity\EntryComment {#1683
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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date: 2023-11-23 16:39:47.0 +01:00
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17 |
DENIED
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Show voter details
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DENIED
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App\Entity\EntryComment {#1548
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About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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Show voter details
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19 |
DENIED
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edit
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App\Entity\EntryComment {#1548
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+body: """
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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Show voter details
|
20 |
DENIED
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App\Entity\EntryComment {#1548
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+body: """
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.\n
\n
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.\n
\n
This is because:\n
\n
A = π r²\n
\n
R = ρ × L / A\n
\n
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
"""
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0 => App\Entity\EntryComment {#1548}
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I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t *think* they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.\n
\n
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.\n
\n
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
"""
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