Does the perpetual license cover upcoming Vray versions?

Thinking about getting a Vray license. Will the perpetual license also cover whatever Vray version comes after the current Vray for Rhino version? Currently it is called “Next” I think, which is equivalent to version 4, right? Will the perpetual license still be valid for a future Vray(5) for Rhino?

Thanks.

No it does not - ‘perpetual’ licenses only include the current version, but you can use that for an unlimited amount of time.

Only the timed ‘subscription’ licenses (yearly, monthly) include the latest available version.

There will most likely be a reduced-price upgrade though, for a limited period of time after introduction.

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OK, thanks a lot for clarifying that. The price for the perpetual license would have been too good if it would cover not only the current version. Would still be worth considering if the vray version release cycle is longer than 2 years.

Hello, everyone,

А perpetual V-Ray license includes all service packs within the same version. Purchasing a V-Ray Next for Rhino license will not make you eligible for a free upgrade to V-Ray 5.
If you choose to use annual/monthly V-Ray license you will always be eligible to use the latest V-Ray version and will not worry about upgrade costs
On a side note, if you obtain a perpetual V-Ray license while there is an active beta for that product you will be eligible for a free upgrade once the new version is officially released.

For further info, please reach out to sales@chaosgroup.com
The dedicated V-Ray forums may also be a more suitable place for such discussions: https://forums.chaosgroup.com/

Kind regards,
Peter Chaushev
V-Ray for Rhino QA Specialsit
www.chaosgroup.com

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Thanks for further detailing this.

Btw, can you give a rough estimate on when to expect vray5 for rhino?

No ETA at this point, I’m afraid. The only info I can share right now is that the dev team is working hard on it. Once a public beta is announced, more info will be available.

Kind regards,
Peter

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OK, they are on it, that’s already a satisfying answer for me, thanks.