Tim B II

From my perspective, there's no such thing as a 'normal' war. Given the relative value of each side to each other, I can't see any material reason for them to go to war and expend so many resources on harming each other. Are you talking about combat as a negotiation tactic for trade?

@JanDorniak yes I have noticed that, and I've made the assumption in my answer that people either know what RSA tokens are and have an understanding of IT security models, or have read the attached link. To be frank, I've assumed it was more obvious than it appears to be.

@Taxi4Dave the randomness of the code generation isn't an issue; the code has already been generated at the time it's valid. It's the display on the RSA token that hasn't occurred yet. As such, even the most 'random' form of code selection is perfectly predictable because the selection occurs before the code is valid, but is only valid before it's displayed on the token.

@JanDorniak the way these encryption dongles work is that the code is only valid while it's displayed on the screen; once the new one comes along, it expires. It's about authentication, not decryption. It's just a way of proving you're you. If you try to do the same thing with the future key by just waiting 30 secs, the key expires and you can't get in. The passcodes from these devices are usually combined with a PIN so that the authentication is a mix of something you know and something you have, but the decryption key will be managed purely by the IT infrastructure; you'll never know it.

@HAHarvey this is a good point, and the above solution isn't foolproof by any means. But it does combine 'something you know' with 'something you have' in a manner that's not possible without this kind of skillset. As with every solution, foresight should be part of your security model, not the entirety of it.

@AlexP so... Wereduke? Wereactor, even Werequeen. I'm really liking where this is going. :) But, thanks for the clarification as it makes better sense to me than my other readings on the topic.

@ArkensteinXII many thanks for the heads up.

+1 for the title of the question alone...

And I upvoted

looks great, I just made a very minor edit to make it look more mathmatical...

Which, unless I've missed my guess, is exactly what you were trying to do...

That way, you're not actually changing how momentum works; you're just putting a proportional upper bound on it based on mass.

Sounds good to me.

(I might also recommend you convert to SI units as this being a science question will attract that comment)

The p_(max) will make sense to the math heads; it means maximum possible P

(with the units in there)

so p=Mv, p(max)= M^2 x 100

I think it's commonly p

But it's good

that

I like it. That makes sense to me. But, I'd recommend putting the equation in to explain it as well as some people like me will benefit from taht

You're very welcome

I'll keep an eye on it, I plan to upvote the question because it's a good one, just needs a bit of clarity. Hope I was helpful

Right-O.

Take a look at what mass^2 x 100 will do for your various projectiles as an upper bound, and then adjust the 100 constant to taste

The thing is, it's a place to start.

Good point

So, you're saying that if the momentum calculation gives a value that exceeds mass^2 x 100, then it's capped at that level

I think you need two equations. The first, momentum, stays the same as it is now, but the new equation is simply an upper bound test

I'm thinking your new momentum equation looks like Momentum = min(mass x v, mass^2 x100)

Hmm.

Well no, because momentum IS mass times velocity. Your previous equation was invalidated by the fact that you were adding units of mass to velocity, which is only speed and direction. It's actually your first equation that's invalidated because of unit crossover.

Or, you can still put in the M^2 x 100 for momentum if you like, so that heavier things can still go faster and have it count

Just call that a momentum equation rather than a velocity equation

Pretty much

Well, it might be a good idea

If you multiply a velocity with a mass, you always get a momentum

The unit types actually

And it's not the math you complicated - it's the value types

I'm just trying to help

No it's all good.

Basically, dealing with the upper bound equation of M^2 x 100 means that you're not automatically applying that much momentum, you're still using it as a cap

Ah. Well if you're looking to nerf impacts based on speed, this still works I think. I'd have to do some more math, but it could work

What you're really trying to do (if I understand correctly) is limit the momentum of an impact by proportionally limiting the upper bound based on the mass and a theoretical speed limit

Well, yes I think so. I was struggling because you're talking velocity when in fact the problem (to me at least) is a momentum one.

That still meets the spirit of your question, makes it a bit clearer in terms of the difference between momentum and upper bound of momentum

Why not say that the upper limit of momentum is mass^2 x 100?

Hmm. More thought required on my part for this

Right. So you're saying that this only provides an upper bound, so to speak?

Much simpler