Zheng Xuan’s target geometries, lateral view, with targets (correctly) set at the appropriate distances (note that the y-axis is expanded 300% for the sake of legibility). Point a represents the archer’s eye, and b the point directly beneath his eye on the ground. Points c and f represent the upper and lower edges of the target, d and e the upper and lower edges of the swan, and g the point directly beneath each target on the ground. Line AY″ runs parallel to the ground through the archer’s eye, intersecting the plane of the respective targets at y, y′, and y″. Line CX and CX′ run parallel to the ground from the upper edge of the forward target to the plane of the target behind it. In racoon-dog paces, BG =AY = 50, BG′ = AY′ = 70, BG″ = AY″ = 90, and G′G″ = CX = C′X′ = YY′ = Y′Y″ = 20. As per problem 9.22 of the Nine Chapters of Mathematical Procedures, we assume AB = YG = Y′G′ = Y″G″ = 7 chi. Zheng Xuan gives CG = 19.2 chi, so CY = XY′ = 19.2 – 7 = 12.2 chi. ACY and CE′X are similar triangles, where E′X/CX = CY/AY, so E′X = 20 × 12.2/50 = 4 ²²/₂₅ chi, and E′G′ = 19 ¹/₅ + 4 ²²/₂₅ = 24 ²/₂₅ chi. As per fig. 2, E′C′ = 13 ⅓ chi, and E′F′ = 8 ⅔ chi, so C′G′ = 24 ²/₂₅ + 13 ⅓ = 37 ³¹/₇₅ chi, and the distance F′G′ from the medium-range target to the ground is 24 ²/₂₅ – 8 ⅔ = 15 ⁵¹/₇₅ chi. For the long-range target, X′Y″ = C′Y′ = 37 ³¹/₇₅ – 7 = 30 ³¹/₇₅ chi, AY′ = 70 racoon-dog paces, C′X′ = 20 racoon-dog paces, and since AC′Y′ and C′E″X′ are similar triangles, E″X′/C′X′ = C′Y′/AY′, so E″X′ = 20 × 30 ³¹/₇₅ ÷ 70 = 8 ³⁶²/₅₂₅ chi, and E″G″ = 37 ³¹/₇₅ + 8 ³⁶²/₅₂₅ = 46 ⁵⁴/₅₂₅ chi. As per fig. 2, E″C″ = 16 chi, and E″F″ = 10 chi, so the distance F″G″ from the long-range target to the ground is 46 ⁵⁴/₅₂₅ – 10 = 36 ⁵⁴/₅₂₅ chi. Q.E.D.