Is Nikon 17-55 worth buying?

First of all, it is certain that the Nikon 17-55 gold ring lens is a level higher than the Tamron lens you use. It has better imaging and feel, and the weight is naturally "better". If you just want to keep the same angle of view as before when using your camera, and don't plan to try different angles of view, then there is no better choice than the Nikon 17-55. After purchasing 17-55, B005 can only be disposed of as second-hand, because it has no existing value.

If what I said in the previous sentence is right, you don’t need to read further. . . If you also want to try a slightly different perspective, you can continue watching^0^. . .

Secondly, my personal opinion: The purchase value of Nikon 17-55 F2.8 is not high! 1. The image quality is not as good as the legend says. Its edge image quality is not high, not even better than the Sima 18-50 and Tamron 17-50 (A16). You can refer to this foreign professional lens testing website. I am not saying this casually. 2. A DX half-frame lens costs 9,00, which is too expensive, and there is no room for upgrade (although DX lenses can be used on FX machines, the angle of view will not become a full-frame angle of view, which is a waste of the body). Maybe in a year or two the price of a D700-level full-frame machine will be the same as the current D300S. Do you plan to use the D300S for a lifetime? If you buy a full-frame camera two years later, will you want to cry when you see this 17-55?

Listening to what you say, there should be very few machines and lenses, but the income is pretty good. If you can spend so much money on a 17-55, why not spend a little more on a 24-70 F2.8G? It is a true full-frame camera with better image quality, but the viewing angle is a little farther. If you use it on a full-frame body, it will be the same as what you are using now. 17-55 is a feeling. And after entering 24-70, the current B005 can be retained for the time being, because it still has some use in wide angle. You can think about the above yourself^0^.