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Confuse the Offense
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<blockquote data-quote="Techster" data-source="post: 442553" data-attributes="member: 360"><p>I was thinking more like a 35 yard throw...but I can accept 30 yards. <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p><p></p><p>The WR breaks at the 40 yard, and the QB completes his drop step at the 23 but steps up to the 22 yard line when the WR makes his break...the optimal time to deliver the ball. That's a 22 yard STRAIGHT LINE distance...except the QB is on the far hash so it's actually a diagonal throw across the field.</p><p></p><p>Using the pythagorean theorem, and solving for the hypotenuse (the diagonal throw), "A" being the 22 yards (or 66ft) between the QBs drop and 40 yard line straight distance where the WR breaks. Distance between college hashes is 40 ft. Distance between far hash and where the WR ended up (exactly on the number) is about 70 (40ft + 30 ft (far hash to sideline is 60ft, so half that)). Now we know "B" equals 70ft.</p><p></p><p>Solving for the hypotenuse which is the square root of A^2+B^2 = square root of (66^2)+(70*2) = 96.21. Divide that by 3 (converting to yards) gives us a little over 32 yards.</p><p></p><p>And thus concludes a GT nerd's post in a sports forum...<img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite27" alt=":whistle:" title="Whistling :whistle:" loading="lazy" data-shortname=":whistle:" /></p></blockquote><p></p>
[QUOTE="Techster, post: 442553, member: 360"] I was thinking more like a 35 yard throw...but I can accept 30 yards. :) The WR breaks at the 40 yard, and the QB completes his drop step at the 23 but steps up to the 22 yard line when the WR makes his break...the optimal time to deliver the ball. That's a 22 yard STRAIGHT LINE distance...except the QB is on the far hash so it's actually a diagonal throw across the field. Using the pythagorean theorem, and solving for the hypotenuse (the diagonal throw), "A" being the 22 yards (or 66ft) between the QBs drop and 40 yard line straight distance where the WR breaks. Distance between college hashes is 40 ft. Distance between far hash and where the WR ended up (exactly on the number) is about 70 (40ft + 30 ft (far hash to sideline is 60ft, so half that)). Now we know "B" equals 70ft. Solving for the hypotenuse which is the square root of A^2+B^2 = square root of (66^2)+(70*2) = 96.21. Divide that by 3 (converting to yards) gives us a little over 32 yards. And thus concludes a GT nerd's post in a sports forum...:whistle: [/QUOTE]
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