Uhm not really there is a difference between what it will reserve versus use, it appears as if it's using all of it, but it's not.
Jensen stated the 3060 has double the rtx performance of the 2080ti if that helps what it means though and how that translates to resolution and frames we will see may be a different story, either way people jump the gun, synthetics are 1 thing but we need to wait to see actual performance numbers of the games themselves to see if the price is warranted, currently the bottleneck is the monitor market anyways.I'm really interested in seeing what ray-tracing performance these cards will have now that the tech will be out of "beta". (and I would like to thank the RTX 2000 series beta testers for their sacrifice)
Depends on how NVcache is used as well.
I would probably say yes Cyberpunk will be the poster child, to show how and why you should upgrade, for some it will justify the price, others well according to comments across the net many already made their mind up beforehand.Oh Cyberpunk 2077 will fully utilize even the RTX 3090 alright.
There is nothing such as enough performance
Now the only question is can gamers afford those extravagant GPU...and looking at Nvidia financial situation the answer is probably yes.
Want to bet on it?
The forum censored my post, I feel violated
I believe the actual card has a 220v 3-phase plug, to get around the 120v 1500 watt limit.
It would have been more funny with some context, so here you go...
It seems Ratchet & Clank has a 60 FPS mode with upscaling and a native 4K 30FPS mode. It's good that they are at least they offering the option for 60FPS considering how good it looks.
Devil is in the details. I predict both that initial supply will sell out, but also that total quantity will be a tiny fraction of the total video card market, meaning you could both be right.
TDP is about heat transfer and this will always be less than the electrical power consumed (one can’t make energy from nowhere), hence TGP is more useful for determining PSU requirements - it’s also directly measurable via onboard sensors on the card’s +12V lines.
Isn't the correct statement that TDP will always be equal to the electrical power consumed? If a card is drawing 250 watts at a moment in time, it is generating 250 watts of waste heat.
No, because some of that drawn energy gets stored in magnetic fields (in the inductors) and used to transfer charge across electric potentials (in the transistors and diodes). The large majority of it does end up as heat though.
Both PL1 and PL2 are 'TDP' values - Intel uses the term to mean power consumption, rather than heat transfer (something that can be verified with their own power gadget).
True, but its not stored there permanently. An RLC circuit continually oscillates ... at any given moment, it may be storing or releasing energy, but integrated over any significant time period, the average is zero.
When you transfer charge across a transistor or diode junction, you do so against a resistance. That generates an amount of heat exactly equal to the work done (the energy consumed) in transferring the charge.
The arithmetic mean for the component may be average, but the environmental losses aren't (I.e. e/m fields aren't fully constrained). They're not significant though, of course.
I obviously over-simplified matters, because with MOSFETs, you also have a number of capacitive losses. But even if they, and the resistive losses, didn't exist, work is still done is transferring charge. Hence, the total energy required is the potential change + resistive + capacitive losses, but it's only the resistive loss that really becomes heat. But of course, even this is far too simplistic a view. This paper provides far more depth on the matter (and it's a very good read):
