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So.. like, bowling ball snooker on a football size field.
...
I'd watch it.
[link]
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How are you holding the cue? |
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Surely you've seen an Olympic size pool before? |
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Bowling balls won't hold together under the forces needed to move them that far with enough energy to spare to move other balls. And you're going to need one heck of a cue. |
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Step 1: Scaling the Pool Table and Balls |
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Standard Pool Table Dimensions: A 9-foot pool table is approximately 2.54 m long and 1.27 m wide. The cue ball diameter is 0.05715 m (2.25 inches). |
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Football Pitch Dimensions: A standard FIFA football pitch is 105 m long and 68 m wide. |
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Scaling Factor: The linear scaling factor is based on the length ratio:
Scaling factor = Football pitch length / Pool table length = 105 / 2.54 = approximately 41.3386
Let's compute this precisely:
105 divided by 2.54 = 41.338582677
So, the scaling factor is approximately 41.34 (rounded to two decimal places for simplicity, but I'll carry the precise value for calculations). |
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Scaled cue ball diameter:
0.05715 * 41.338582677 = approximately 2.3625 m
This confirms the scaled ball diameter is about 2.36 m. |
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Mass Scaling: Mass scales with the cube of the linear scaling factor (m proportional to L cubed). A standard cue ball has a mass of approximately 0.17 kg. The scaling factor cubed is:
(41.338582677) cubed = 41.338582677 * 41.338582677 * 41.338582677
First, compute 41.338582677 squared:
41.338582677 * 41.338582677 = approximately 1708.87962766
Then:
1708.87962766 * 41.338582677 = approximately 70642.706093
So, the scaled mass is:
m_scaled = 0.17 * 70642.706093 = approximately 12009.2600358 kg
This matches the previous calculation of about 12,009 kg, confirming the mass is correct. |
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Step 2: Force Required to Push the Balls
To maintain the same dynamics (e.g., same number of bounces), we assume the scaled-up ball moves at the same absolute velocity as the original cue ball (e.g., 2 m/s for a typical pool shot) to preserve dynamic similarity. The force depends on the impulse required to achieve this velocity. |
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Impulse for Original Cue Ball: m = 0.17 kg, v = 2 m/s Impulse = m * v = 0.17 * 2 = 0.34 kg m/s If the cue contacts the ball for 0.001 s (typical for a pool shot): F = Impulse / delta t = 0.34 / 0.001 = 340 N This is reasonable for a human-powered cue shot.
Impulse for Scaled-Up Ball: m_scaled = 12009.2600358 kg, v = 2 m/s Impulse_scaled = 12009.2600358 * 2 = 24018.5200716 kg m/s Using the same contact time (0.001 s): F_scaled = 24018.5200716 / 0.001 = 24018520.0716 N = approximately 24.02 MN This is a very large force, as noted previously. To make it more practical, assume a longer contact time of 0.1 s (e.g., a mechanical device like a hydraulic ram): F_scaled = 24018.5200716 / 0.1 = 240185.200716 N = approximately 240.19 kN The previous calculation gave 240,180 N (about 240 kN), which is very close. The slight difference comes from rounding the mass (12,009 kg vs. 12,009.26 kg). Using the precise mass, the force is 240,185.20 N, confirming the order of magnitude is correct.
Friction Check: For a bowling-alley-like surface with a coefficient of friction mu = 0.05: F_friction = mu * m * g = 0.05 * 12009.2600358 * 9.81 = approximately 5892.53939376 N = about 5.89 kN The previous value was about 5,893 N, which is consistent given rounding. |
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Step 3: Material Strength Check |
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Pressure from Ball Weight: Each ball (12,009.26 kg) exerts a force: F = m * g = 12009.2600358 * 9.81 = approximately 117810.840951 N Assuming a contact area of 0.01 m squared: P = F / A = 117810.840951 / 0.01 = approximately 11.781 MPa This matches the previous about 11.8 MPa, confirming that materials like high-strength steel or reinforced concrete are suitable. |
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Step 4: Verification of Assumptions |
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Dynamic Similarity: The assumption that velocity remains the same (2 m/s) ensures the number of bounces scales appropriately, as the coefficient of restitution (COR) and friction are kept similar. The COR of 0.8 to 0.9 for rails and low friction (mu approximately 0.04 to 0.06) are consistent with a lubricated surface.
Contact Time: The choice of 0.1 s for the scaled-up force is more realistic than 0.001 s, reducing the force to a manageable level for heavy machinery. |
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Corrections and Clarifications
The math checks out with minor differences due to rounding (e.g., 240,185 N vs. 240,180 N). The scaling factor, mass, impulse, and force calculations are consistent. The material requirements (steel base, PTFE-coated surface, rubber rails) remain valid based on the pressure and friction calculations. |
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Force Required: Approximately 240,185 N (about 240.19 kN) to impart a 2 m/s velocity to a 12,009.26 kg ball over 0.1 s, ensuring the same bounce dynamics as a standard pool table.
Material: High-strength steel base with a polished steel or ceramic surface coated with PTFE (mu approximately 0.04) for low friction, and tempered steel or hard rubber rails for a high COR (0.8 to 0.9). |
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That you can math that fast makes you one of my heroes. |
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I can't math. I mean... I might be able to learn how to math... but since I'm nearing 60 it seems unlikely that I will get to do so before I die. |
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I don't know how to de-cypher what you've written. It kinda sucks. |
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I have a complete set of eight-ball golf balls and wonder how much interest there will be in playing eight-ball golf on a regulation eight-ball sized table for money. |
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I'll let you guys know when I find out. It's what inspired this posting. I figured golf balls on a table? cool, what about bowling balls on a football field? |
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Oh come on, that was funny and you know it! |
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I can math, but I can't math that quickly. That was Grok. Also I cannot physics. And mathing is a lot easier than you think, just don't be intimidated by it. Especially don't be intimidated by the mathematical shorthand notation, which makes things unnecessarily confusing. Rather approach things from the bottom asking yourself what needs to be solved and for that what needs to be multiplied and divided and so on. |
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The "how many apples can fit on uranus" was me, and I bet if you approach that problem you'll be able to figure it out for a fun project, just remember that the packing density of cylinders is 2/3 |
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