Can 1 dmg be roughly estimated in Joules? [closed]

Related to this question: How many Joules does strength values ten through twenty estimate to?.

The same players are curious if the amount of force required to do 1 dmg can be roughly estimated in Joules?

Physics issues:

  • The primary worlds of D&D (Greyhawk, Forgotten Realms, and Dragonlance) have been established as having Earth-similar physics as the default when not overridden by psionics, magic, incarnum, etc..
  • All tests are taking place at sea level and in a gravity normal area (1 atmosphere, and 1G).
  • no special effects are interfering, allowing for normal physics to operate and be measured.
  • average humans are being used to conduct this experiment.
  • no beings were harmed in the process of these experiments.
  • Materials from the hardness / breaking tables are being used as targets, and are of a size and weight to exactly match the values listed on the table.

Though experiment issues:

  • Using the strength values given in the previous question, the Joules of STR bonus could be roughly estimate, however explosive striking force (from a punch or kick) is likely a different value from lifting a weight overhead force.
  • Effects like Control Fire give values of damage for different sizes of fire, and comparing the size of fire with the amount of damage it will do to various objects on the hardness/breaking tables which might serve as a possible yardstick to estimate.
  • HP is a universalized generalization which almost certainly does not equate to real world values of health and/or structural damage capacity of flesh or rock etc, for example, and average first level commoner might have an hp of 2, and a small pebble has 2 hp, but the amount of force required to deal "1 hp" of damage might not be equivalent.

It might not be possible to come up with an estimate value for 1 dmg due to the possibly inconsistent values of hp between different materials, however, please give it a go. Feel free to make whatever assumptions or estimates are necessary to come up with a rough estimate.

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View estimated size of GitLab repository before cloning

I would like to view the total repo size of a project hosted on GitLab without having to clone it.

I was able to find one for HitHub here, but I can’t figure out how to do it on Gitlab

The HitHub example:

$   echo | perl -ne 'print $  1 if m!  ([^/]+/[^/]+?)(?:\.git)?$  !' | xargs -i curl -s -k'{}' | grep size    >>"size": 284, 

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I have installed adsense on my website a bit over a month ago. During the first few days estimated earnings started to update my overall balance (increasing it). My balance was at ca.80usd from having previously earned ad money from a youtube account.

Now, a month later, despite seeing estimated earnings and clicks every day, my balance remains at ~96usd and hasn’t budged since. My estimated earnings over this time is at 66usd. Added on to my balance it should be over 100usd (the min threshhold for being paid out).

Several years ago I have already gotten paid out once before thanks to the youtube earnings. So the bank account that is linked is correct and works. Furthermore, my account is shown as beeing approved (and confirmed).

I have read through the google help documentation, but couldn’t see a reason for my balance not updating. I contacted google using their form, but ofcourse they don’t respond.

Has anyone had this problem and if so did you manage to get paid by google and how?

iOS switch: estimated cost of repurchasing apps

I am looking in to switching from iOS to android. However I have a ton of apps that I have bought in the iOS app store over the past years . I am aware that I will likely have to repurchase them on Android but I would like to have at least an estimate or idea how much this will be. Is there a website or an efficient way to pitch ones iOS app purchases against the android store to see what the estimated cost would be for repurchasing apps? Even if it is just a rough estimate?

In scrum, if tasks are estimated in hours, how to avoid assigning the task in sprint planning?

I’ve been reading up a lot on the subject, and from what I can gather the following statements ‘should’ be true in scrum:

1) Estimate user stories in story points at the product backlog level. This allows abstraction of the estimates, and the team can estimate together without dispute – e.g. one developer may complete the story much quicker than another, but for both, the estimate is 3 story points.

2) You ideally shouldn’t assign tasks during sprint planning. While it helps with indivdual accountability, it reduces team accountability

3) In sprint planning, flesh out the tasks involved for each story, and estimate these in hours.

My question is this, how can you estimate the tasks in hours, yet still not assign the task to a specific developer (points 2 and 3)? Surely the whole point of using story points is to avoid the expert developer saying it’ll take 1 hour, and the junior dev saying it will take 8 hours.

If you are now estimating tasks within sprint planning in hours, how do you reach an estimate which is good for all team members of varying levels of expertise, without assigning the task to the specific dev who’s estimated it?


How to find bilateral confidence interval for an estimated variance?

In constructing a bilateral confidence interval for the variance, is it (n-2) degree of liberty or (n-1) ? I see that in some it’s n-1 and elsewhere n-2. Here is what I mean

Is it this

$ $ \left[ \dfrac {\left( n-1\right) S^{2}_{n}}{\chi_{\left( \alpha /2;n-1\right) }};\dfrac {\left( n-1\right) S^{2}_{n}}{\chi_{\left( 1-\alpha ,2;n-1\right) }}\right] $ $

Or this $ $ \left[ \dfrac {\left( n-2\right) S^{2}_{n}}{\chi_{\left( \alpha /2;n-2\right) }};\dfrac {\left( n-2\right) S^{2}_{n}}{\chi_{\left( 1-\alpha ,2;n-2\right) }}\right] $ $

And Why?

Estimated value of sample variance for normal

Let $ X_1 \ldots X_n$ be iid $ N(0, \sigma^2)$ observations.

I’m told that the correct answer is $ E[S^2] = \sigma^2$ .

If I use the given value of mean in my sample variance calculation then I get the correct answer, but if I use $ \overline{X}$ instead then I get a different answer. Why is that? It could be because my calculations are wrong so I’ve included them here

$ E[S^2] = \Sigma \frac{(X_i – \overline{X})^2}{n} = \frac{n}{n}E[(X_i – \overline{X})^2]$ for any $ i$ . Let $ X_i$ just be $ X$ , then

$ = E[X^2 + \overline{X}^2 – 2\overline{X}{X}]$ $ = \sigma^2 + 0 + \frac{\sigma^2}{n} + 0 – 2E[\overline{X}{X}]$

This comes from $ \overline{X} \sim N(0, \frac{\sigma^2}{n})$

Now consider $ E[\overline{X}{X}] = \frac{1}{n} (E[X^2] + (n-1)E[XX’]) = \frac{1}{n}E[X^2] = \frac{\sigma^2 + 0}{n} = \frac{\sigma^2}{n}$

So $ E[S^2] = \sigma^2 + \frac{sigma^2}{n} – 2\frac{sigma^2}{n} = \frac{n-1}{n}\sigma^2$