Originally posted by h1a8
You basically ignored what I said. I just proved that I didn't. Reread the part about the beam slicing through all materials (aluminum and window) at equal speed. An all-aluminum makeup would have resulted in the same duration of time.Imagine a sweeping laser vaporizing a series of 10 lined-up blocks, 8 made of aluminum and 2 made of acrylic glass. The beam sweeps through the aluminum blocks at the same speed as the acrylic blocks.
Therefore, replacing all the blocks with aluminum blocks means the beam would vaporize all 10 blocks in the same amount of time as the previous mixed set of 10.
But the energy required would not be the same, to laser through 8blocks of aluminium+ 2 blocks of acrylic, Vs 10 blocks of aluminium (your initial assertion).
So your original post about his HV being able to go up to 20,000 degrees C is grossly inflated, as I said.
Originally posted by DarkSaint85
But the energy required would not be the same, to laser through 8blocks of aluminium+ 2 blocks of acrylic, Vs 10 blocks of aluminium (your initial assertion).So your original post about his HV being able to go up to 20,000 degrees C is grossly inflated, as I said.
The logic might have been unclear. Let me simplify it further.
Line up five blocks in a row from left to right on a flat surface. Have a laser sweep at a constant speed from left to right, destroying them one at a time over a total of 5 seconds.
What's the rate at which the laser, moving at that speed, CAN destroy blocks lined up in a row? It's 1 block per second, right?
Now, repeat the process but remove the middle block.
What's the rate at which the laser, moving at that speed, CAN destroy the remaining blocks? Is it 0.8 blocks per second or still 1 block per second?
Did removing the middle block change the rate of the CAN ?
If you say yes, then you're either mistaken or not understanding the concept. Removing a block does not change the rate at which the laser CAN destroy blocks.
If this is still unclear, just calculate the destruction of the first block in the first second to achieve the same rate.
Originally posted by DarkSaint85
But your entire post on the temperature was predicated on the HV melting X volume of aluminium in Y time.My point is that the X volume of aluminium is too much. Thus, the temperatures reached is grossly inflated.
Bad logic (your conclusion does not follow from your premises).
Aluminum is too much as well as the time.
I counted the time where the HV was hitting the window.
I proven that the rate stays the same.
Fill in the window with aluminum and count the time
Or
Keep the window and subtract the extra time to achieve the same result.
Originally posted by h1a8
Bad logic (your conclusion does not follow from your premises).Aluminum is too much as well as the time.
I counted the time where the HV was hitting the window.
I proven that the rate stays the same.Fill in the window with aluminum and count the time
Or
Keep the window and subtract the extra time to achieve the same result.
So the rate staying the same means nothing. Walk us through your calculations, like you did in the other thread. Start with assumptions etc. This is the only way to help you see your mistake.
Originally posted by DarkSaint85
But the amount of energy required to melt acrylic is much less than that of aluminium.So the rate staying the same means nothing. Walk us through your calculations, like you did in the other thread. Start with assumptions etc. This is the only way to help you see your mistake.
Lack of understanding.
The HV was melting the aluminum at a specific rate.
The energy involved is irrelevant.
It's all about the rate.
I never calculated the total energy at all.
Where do you get energy from?
Originally posted by DarkSaint85
Well, pedantry eh?Ok, let's get back to basics - show your workings out, to arrive at the HV being at a temperature of 20,000 degrees C. All assumptions.
I have completed the calculations, but before presenting them, let's establish some fundamental principles.
Imagine 5 blocks, equally spaced in a row. A laser sweeps at a constant rate of 1 block per second, destroying all 5 blocks in 5 seconds. Thus, the laser can destroy 1 block per second.
Now, if the middle block is removed, the laser will still destroy the existing 4 blocks in 5 seconds. We know that if the middle block were present, it would also be destroyed, maintaining the laser's destroy rate of 1 block per second. The missing block does not change the laser's capability, only the result of its action.
Applying this principle to Homelander's feat: Even though part of the plane was made of windows, if it were all aluminum, the heat vision would still sweep and melt the plane at the same rate.
Therefore, it is reasonable to calculate the feat as if the entire plane were aluminum, as this does not affect the heat vision's capability.
With that established, I can post the calculation upon command.
Originally posted by h1a8
I have completed the calculations, but before presenting them, let's establish some fundamental principles.Imagine 5 blocks, equally spaced in a row. A laser sweeps at a constant rate of 1 block per second, destroying all 5 blocks in 5 seconds. Thus, the laser can destroy 1 block per second.
Now, if the middle block is removed, the laser will still destroy the existing 4 blocks in 5 seconds. We know that if the middle block were present, it would also be destroyed, maintaining the laser's destroy rate of 1 block per second. The missing block does not change the laser's capability, only the result of its action.
Applying this principle to Homelander's feat: Even though part of the plane was made of windows, if it were all aluminum, the heat vision would still sweep and melt the plane at the same rate.
Therefore, it is reasonable to calculate the feat as if the entire plane were aluminum, as this does not affect the heat vision's capability.
With that established, I can post the calculation upon command.
So start with the assumptions, then show your working. I see the first assumption you are using is that it is continuous aluminium (i.e. assumption that there are no windows etc), based on the above reasoning.
Originally posted by DarkSaint85
Cool.So start with the assumptions, then show your working. I see the first assumption you are using is that it is continuous aluminium (i.e. assumption that there are no windows etc), based on the above reasoning.
Homelander's HV feat calculation
or more cleaner
Originally posted by DarkSaint85
Yeah, straightaway I have pinpointed where your calculations are wrong. Your heat transfer coefficient is wildly over optimistic, seeing as it assumes a mass and surface area of aluminium that assumes no windows.As I said from the start.
Originally posted by DarkSaint85😄
Cool.So start with the assumptions, then show your working. I see the first assumption you are using is that it is continuous aluminium (i.e. assumption that there are no windows etc), based on the above reasoning.
Also, in this case, mass is directly proportion to surface area. Reducing the mass (exclude the windows) also reduces the surface area by the SAME FACTOR.
One more thing. Man made lasers have heated aluminum to over 3 million degrees. 20,000 degree Celsius is nothing compared to that. So I don't see how 20000 is overly optimistic. It's really a gross lowball estimate. I didn't even include the mass of the stringers and formers (which add significant more mass of aluminum).
He cut the plane at 60 degrees. That means the length of the cut is significantly bigger than the height of the plane. Again more mass.
The thickness of a plane's skin can be over 2 mm and about 1.4mm average for smaller planes. I used 1 mm. That's 40 percent less mass I used.
Lastly I assumed he just heated the aluminum to 660 (the beginning of being melted) when in reality it could have been heated to over 1000 degrees.
I took liberty to lowball every number instead of at least using averages.