Correction on Cloud Calculations page

[Floxy178]

So recently I was going through Cloud Calculations page and noticed this:

In the case of such an omnidirectional expansion one could use the formula "kinetic energy = 0.25 * cloud mass * (Speed of cloud movement)^2" to account for the different speeds involved.

For pulling in clouds omnidirectional towards a center point, or dispersing them from a center point, the formula "kinetic energy = 1/12 * cloud mass * (Speed of cloud movement)^2" can be used.

However, "pulling in clouds omnidirectional towards a center point" should also be 0.25 * m * (max speed)^2, not 1/12. It's same as first case (expansion from center point) but reversed. Both of them have same speed distribution(0->v from center to edge, only direction is different), so same parts of cloud for both cases will have equal speed, and thus equal KE.

 

[KLOL506]

So only the dispersal will have the 1/12 method I presume?

 

[Floxy178]

So only the dispersal will have the 1/12 method I presume?

Yes.

 

[Damage3245]

On this topic, I think the Cloud Calculations K.E. section should also be updated to explain why 1/12 * cloud mass * (Speed of cloud movement)^2 would be used. Right now it just says that it can without explaining why it is correct to use.

 

[Floxy178]

On this topic, I think the Cloud Calculations K.E. section should also be updated to explain why 1/12 * cloud mass * (Speed of cloud movement)^2 would be used. Right now it just says that it can without explaining why it is correct to use.

I can write it if you want.

I myself was writing one for relativistic cases.

 

[Floxy178]

On this topic, I think the Cloud Calculations K.E. section should also be updated to explain why 1/12 * cloud mass * (Speed of cloud movement)^2 would be used. Right now it just says that it can without explaining why it is correct to use.

I searched a little and found it. That's thread where it was added.

However formula provided by DT uses different velocity distribution, like he assumes maximum speed at center and 0 for edge. Opposite should be used, just like for omnidirectional expansion case.

In the formula, we should replace (r-x)/t by just x/t, which will result in 1/4 mv^2. (r-x)/t should be used for dispersing case where center has maximum speed and edge has 0.

 

[DontTalkDT]

No, pulling clouds in from the edge is the time reversal of dispersing them from the center, not the time reversal of spreading them from the center.

To illustrate:

1. Spreading from the center: First clouds in the center, in the end clouds everywhere.
2. Compressing towards the center: First clouds everywhere, then clouds only in the center.
3. Dispersing them from a center point: First clouds everywhere, then clouds all at the edge.
4. Pulling clouds from the edge: First clouds at the edge, then clouds everywhere.

1 and 2 are reversals of each other, and 3 and 4 are reversals of each other. 4 is not the reversal of 1 (else all would be the same).
The important difference, formula derivation wise, is which clouds have the maximum velocity here.
In case 1, the maximum velocity is reached by the clouds that in the end are in the edge (i.e. the biggest circle). In case 4, the highest velocity is reach by the clouds that end up in the center (i.e. the smallest circle). Which illustrates quite well why the latter formula returns the lower results: It requires the biggest KE for the clouds of which the least are present, while the former formula requires the biggest KE for the clouds of which the most are present.

 

[Floxy178]

No, pulling clouds in from the edge is the time reversal of dispersing them from the center, not the time reversal of spreading them from the center.

To illustrate:

1. Spreading from the center: First clouds in the center, in the end clouds everywhere.
2. Compressing towards the center: First clouds everywhere, then clouds only in the center.
3. Dispersing them from a center point: First clouds everywhere, then clouds all at the edge.
4. Pulling clouds from the edge: First clouds at the edge, then clouds everywhere.

1 and 2 are reversals of each other, and 3 and 4 are reversals of each other. 4 is not the reversal of 1 (else all would be the same).
The important difference, formula derivation wise, is which clouds have the maximum velocity here.
In case 1, the maximum velocity is reached by the clouds that in the end are in the edge (i.e. the biggest circle). In case 4, the highest velocity is reach by the clouds that end up in the center (i.e. the smallest circle). Which illustrates quite well why the latter formula returns the lower results: It requires the biggest KE for the clouds of which the least are present, while the former formula requires the biggest KE for the clouds of which the most are present.

Point 2 is exactly how I imagined the description, because it doesn't mention cloud's state before pull(so I guess we need to reword it?). In the thread of 0.25 one pull towards center was described as "opposite scenario".

Not really sure that pulling clouds initally compressed at edge and forming a disk/cylinder would be more common way of interpretation of "pulling clouds to center point from every direction" intead of regular cloud being compressed to center.

What do you think about mentioning 1/12 one is pull from edge or simply adding that formulas can be applied reverse of situations too?

 

[DontTalkDT]

It might be best to get a bit more detailed on all four cases to avoid confusion.

 

[Floxy178]

In the case of such an omnidirectional expansion, or the opposite, compression of clouds in center point, one could use the formula "kinetic energy = 0.25 * cloud mass * (Speed of cloud movement)^2" to account for the different speeds involved.

For pulling in clouds omnidirectional towards a center point from the edge, or the opposite, dispersing them from a center point, the formula "kinetic energy = 1/12 * cloud mass * (Speed of cloud movement)^2" can be used.

What do you think? Imo mentioning them being "opposite" would be better than explaining those scenarios with extra detail individually.

 

[Floxy178]

Btw I made visuals for them.

Case 1&2

Case 3&4

Not that good looking but better than nothing I guess.

 

[DontTalkDT]

The visuals look pretty good as far as I am concerned. Could you separate them into the 4 cases (not having 1&2 in the same video is better IMO) and write a short explanation of what in the visual represents what?

What do you think? Imo mentioning them being "opposite" would be better than explaining those scenarios with extra detail individually.

That sounds ok to me.

 

[Floxy178]

The visuals look pretty good as far as I am concerned. Could you separate them into the 4 cases (not having 1&2 in the same video is better IMO) and write a short explanation of what in the visual represents what?

Sure.

 

[Floxy178]

Bump

 

[Floxy178]

Bump

 

[DontTalkDT]

Great, as far as I am concerned, this can be applied.

In the case of such an omnidirectional expansion, or the opposite, compression of clouds in center point, one could use the formula "kinetic energy = 0.25 * cloud mass * (Speed of cloud movement)^2" to account for the different speeds involved.

For pulling in clouds omnidirectional towards a center point from the edge, or the opposite, dispersing them from a center point, the formula "kinetic energy = 1/12 * cloud mass * (Speed of cloud movement)^2" can be used.

What do you think? Imo mentioning them being "opposite" would be better than explaining those scenarios with extra detail individually.

@KLOL506 @Damage3245 You guys fine with applying that change in wording with the corresponding illustrations linked?

 

[KLOL506]

Looks alright

 

[Floxy178]

Thanks for evaluation

I suppose 2 votes are enough?

 

[DontTalkDT]

Did the thing.
Does that work or did I make some mistake?

 

[Damage3245]

Did the thing.
Does that work or did I make some mistake?

Looks good, but can an additional explanation be added onto the page? It says:

the formula "kinetic energy = 1/12 * cloud mass * (Speed of cloud movement)2 " can be used.

But it doesn't say why this should be used, why it'd be correct.

 

[Floxy178]

Did the thing.
Does that work or did I make some mistake?

It looks good to me.

Looks good, but can an additional explanation be added onto the page? It says:



But it doesn't say why this should be used, why it'd be correct.

Maybe we can link the thread where it was accepted.

 

[Floxy178]

Maybe we can link the thread where it was accepted.

Bump?

 

[Damage3245]

@DontTalkDT Could you help with the problem up above?

 

[KLOL506]

bump

 

[Kaydee1648]

bump