Discover and read the best of Twitter Threads about #Volta
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Let's take a tour along the coast of eastern #Ghana, #Togo and #Benin in images taken from #ISS, 11Dec2019. First stretch is eastern Ghana. Images ISS061-E-76460-76489. We start at #OldNigo, hi-res eol.jsc.nasa.gov/DatabaseImages… @DaveAtCOGS 1/6 
Next stop is at #SongorLagoon. Images ISS061-E-76469-76473. Hi-res eol.jsc.nasa.gov/DatabaseImages… @DaveAtCOGS 2/6
We then take a look at the #Volta River entering the Atlantic Ocean (and the Gulf of Guinea in particular). Hi-res eol.jsc.nasa.gov/DatabaseImages… @DaveAtCOGS 3/6
Back of the envelope calculation:
RTX 2080Ti: 10GRay/s @ 616GB/s mem bandwidth = 61 bytes/Ray
1 triangle, 3x 32 bit float3 vertices: 48 bytes
61 - 48 = 13 bytes left for BVH traversal
That would be under an ideal BVH that requires only 1 ray triangle intersection/ray
RTX 2080Ti: 10GRay/s @ 616GB/s mem bandwidth = 61 bytes/Ray
1 triangle, 3x 32 bit float3 vertices: 48 bytes
61 - 48 = 13 bytes left for BVH traversal
That would be under an ideal BVH that requires only 1 ray triangle intersection/ray
Compressed wide BVH (research.nvidia.com/sites/default/…) requires 80 Bytes per BVH node. A balanced BVH8 over 1 million triangles is 7 level deep, so we're looking at 80 bytes * 7 = 560 bytes of processed data per ray. Times ten gigarays/s = 5.6 TB/s of bandwidth just for BVH traversal.
#Volta #V100 has 12-14TB/s shared memory bandwidth (arxiv.org/pdf/1804.06826…), so 10GRays/s are plausible if most of the data fits in L1 cache/shard mem.
V100 has 80 SMs with 128KB L1/shared mem each, a total of 10MB. 10MB aren't enough to fit a 7 levels deep BVH8.
V100 has 80 SMs with 128KB L1/shared mem each, a total of 10MB. 10MB aren't enough to fit a 7 levels deep BVH8.
