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Model X Price higher than announced?

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While I can't say if that is normally true or not, that's not how Tesla has been talking.
Look at the Cd formula on the Wikipedia page and you will see that what I described is accurate.
They were trying to point out that they made the X "slipperier" but that the larger frontal area more than offsets that. combined with the larger vehicle having more weight, there's no way for the X to be more efficient overall than the S, but that doesn't mean that it isn't more efficient in relation to it's size and weight.
Maybe, but given what we have seen so far of the X body design it is hard for me to understand how the X body is "slipperier" and has a lower Cd than the S.
 
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Look at the Cd formula on the Wikipedia page and you will see that what I described is accurate.

Maybe, but given what we have seen so far of the X body design it is hard for me to understand how the X body is "slipperier" and has a lower Cd than the S.

Basically the Cd is the total drag divided by the frontal area. Multiply it again by the frontal area and you get back the total drag. To compare Cd between the X and the S you need to imagine shrinking the X so that the frontal area is the same as the S and then comparing the shape. Not that it makes much real difference except bragging rights. Total drag is what counts.
 
Maybe, but given what we have seen so far of the X body design it is hard for me to understand how the X body is "slipperier" and has a lower Cd than the S.
I have no reason to believe that Tesla is lying about this, and I suspect they know better than anyone else what they've managed to do with their design. It's likely small things, slight changes to angles around wheel wells and mirrors, little changes around vents, wiper blades and window trims, and don't forget the undercarriage, the S is good, but that doesn't mean there's no room for improvement. Every little bit adds up. I'm sure they learned a lot from the S that they've been able to use to make little improvements, none of which you'd be able to see from fuzzy pictures of the X body posted various places.
 
Thanks Robby. But as I understand it the Cd equation includes A in the denominator, it is not independent of A.
Look at the Cd formula on the Wikipedia page and you will see that what I described is accurate.

Can you be a little more specific? That Wiki page reads to me to agree with what I said, as there is a clear separation between the drag coefficient and the area. It says the force required to overcome drag is 0.5 * air density * drag area * speed^2. Drag area is defined as "the product" of the drag coefficient (Cd) and the area (A). So you need not just Cd, but also A, to multiply them together to get the drag area. Is there something I'm missing here, either on the wiki page or in what you're saying?