WhatIf 0085: "Rocket Golf"
Moderators: Moderators General, Prelates, Magistrates
 Various Varieties
 Posts: 505
 Joined: Tue Mar 04, 2008 7:24 pm UTC
Re: WhatIf 0085: "Rocket Golf"
Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Re: WhatIf 0085: "Rocket Golf"
Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Re: WhatIf 0085: "Rocket Golf"
CasualSax wrote:Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Re: WhatIf 0085: "Rocket Golf"
rmsgrey wrote:CasualSax wrote:Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Are you taking account of the packing fraction? Golf balls don't tessellate very well...
I did not, but I assumed that Mr. Monroe did not either, so the volumes would translate to the same number of golf balls (even though the volume is wrong).
Re: WhatIf 0085: "Rocket Golf"
Randall's last line
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26731
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Yes, he did. That's what the 65% number is here:CasualSax wrote:I did not, but I assumed that Mr. Monroe did not eitherrmsgrey wrote:Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Stripping that equation down to just the mass part, we get 1.616x10^{45}kg for the initial number, which is 3.519x10^{46} 1.62 oz regulationsized golf balls.
For the final number, where each ball is going 140m/s, we get 5.523x10^{18}kg or 1.203x10^{20} balls.
(The number of digits you gave, CasualSax, is incidentally pretty absurd, considering you estimated a single ball to only 2 significant figures. Those extra digits don't give any usable information, they just make it harder to see at a glance how big the numbers are.)
It's not that many tries to hit a ball, though. It's one try for each ball.NOTNOTJON wrote:Randall's last lineUmm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one.over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Re: WhatIf 0085: "Rocket Golf"
gmalivuk wrote:Yes, he did. That's what the 65% number is here:CasualSax wrote:I did not, but I assumed that Mr. Monroe did not eitherrmsgrey wrote:Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Stripping that equation down to just the mass part, we get 1.616x10^{45}kg for the initial number, which is 3.519x10^{46} 1.62 oz regulationsized golf balls.
For the final number, where each ball is going 140m/s, we get 5.523x10^{18}kg or 1.203x10^{20} balls.
(The number of digits you gave, CasualSax, is incidentally pretty absurd, considering you estimated a single ball to only 2 significant figures. Those extra digits don't give any usable information, they just make it harder to see at a glance how big the numbers are.)It's not that many tries to hit a ball, though. It's one try for each ball.NOTNOTJON wrote:Randall's last lineUmm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one.over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
When we are doing napkin math for the sheer sake of absurdity, seeing the full number outright has more comical effect than scientific notation.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26731
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Ah, then I guess my mistake was doing math to answer the question asked...
Re: WhatIf 0085: "Rocket Golf"
NOTNOTJON wrote:Randall's last lineover the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
Since none of those balls were hit from the appropriate tee box (I know of no golf course with a green on earth and a tee box in orbit), none of them are holes in one. A million tries from the tee box would count if one of them went in. Though that's a truly Presidential number of rounds of golf, or a frighteningly permissive mulligan rule.
Re: WhatIf 0085: "Rocket Golf"
flying_kiwi wrote:Olaf Klischat is correct  for almost all configurations discussed, the escape velocity of the ship is greater than the launch velocity of the ball, so it will simply fall back to the surface of the ship after launch, and (assuming collisions are at least partially inelastic and it eventually comes to rest) the net change in momentum of the ship will be zero.
In the final configuration, plugging in the density of the average golf ball I get the launch velocity a little larger than the escape velocity, so the ship WILL move. However, the final velocity is still a good deal less than the launch velocity, which reduces the effectiveness of the rocket, forcing the designer to make it larger  likely to the point you wouldn't be able to launch to escape velocity, and the rocket would fail to change its momentum.
I'm afraid it's even more complicated than that because the ship is a noninertial reference frame. The launch speed doesn't need to exceed the ship's escape velocity because before the first ball drops back to the ship, it will have accelerated away from the ball. This is just based on intuition and I'd like to see someone more mathematically skilled actually write that out as an equation.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26731
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Yeah, but you could still consider it one ball at a time, to get a bound on it. If the launch speed is above the ship's escape velocity, shooting off the golf ball will definitely accelerate the ship. If it's equal or less, then the only way to accelerate is to fire off golfballs fast enough that your own velocity added to the ball's will exceed escape velocity (at a particular distance) before the ball stops.
Later I may actually do the math to find the speed at which it balances out exactly (i.e. where launch speed is equal to the full ship's escape velocity).
Edit: Now that it's later, I'll do the calculation. I'll furthermore do it with the more accurate 63.4% for random close packing and 1.13 kg/L for a golf ball.
If we set the initial equation equal to the radius for a given mass and escape velocity (where the mass is that given by the rocket equation), we need to find v so that
[;
\left(\frac{3}{4\pi}\exp\left(\frac{5300\ m/s}{v}\right)200\ kg \frac{1\ m^3}{.634*1130\ kg}\right)^{\frac{1}{3}}=\frac{2G\exp\left(\frac{5300\ m/s}{v}\right)200\ kg}{v^2}
;]
Then the velocity that works is a bit over 134m/s. Meaning that the last, fastest scenario in the whatif is the only one with any possibility of working at all, even if we ignore all the other matter in the universe.
Edit2: If we use the optimal 74% efficiency packing, the resulting speed is only about a third of a meter per second faster.
Later I may actually do the math to find the speed at which it balances out exactly (i.e. where launch speed is equal to the full ship's escape velocity).
Edit: Now that it's later, I'll do the calculation. I'll furthermore do it with the more accurate 63.4% for random close packing and 1.13 kg/L for a golf ball.
If we set the initial equation equal to the radius for a given mass and escape velocity (where the mass is that given by the rocket equation), we need to find v so that
[;
\left(\frac{3}{4\pi}\exp\left(\frac{5300\ m/s}{v}\right)200\ kg \frac{1\ m^3}{.634*1130\ kg}\right)^{\frac{1}{3}}=\frac{2G\exp\left(\frac{5300\ m/s}{v}\right)200\ kg}{v^2}
;]
Then the velocity that works is a bit over 134m/s. Meaning that the last, fastest scenario in the whatif is the only one with any possibility of working at all, even if we ignore all the other matter in the universe.
Edit2: If we use the optimal 74% efficiency packing, the resulting speed is only about a third of a meter per second faster.

 Posts: 2
 Joined: Sat Mar 01, 2014 6:24 pm UTC
Re: WhatIf 0085: "Rocket Golf"
orbik wrote:flying_kiwi wrote:Olaf Klischat is correct  for almost all configurations discussed, the escape velocity of the ship is greater than the launch velocity of the ball, so it will simply fall back to the surface of the ship after launch, and (assuming collisions are at least partially inelastic and it eventually comes to rest) the net change in momentum of the ship will be zero.
In the final configuration, plugging in the density of the average golf ball I get the launch velocity a little larger than the escape velocity, so the ship WILL move. However, the final velocity is still a good deal less than the launch velocity, which reduces the effectiveness of the rocket, forcing the designer to make it larger  likely to the point you wouldn't be able to launch to escape velocity, and the rocket would fail to change its momentum.
I'm afraid it's even more complicated than that because the ship is a noninertial reference frame. The launch speed doesn't need to exceed the ship's escape velocity because before the first ball drops back to the ship, it will have accelerated away from the ball. This is just based on intuition and I'd like to see someone more mathematically skilled actually write that out as an equation.
I haven't worked it out, but it won't help much  the ball loses most of its momentum immediately after the "launch", and it is during this time that the "equal and opposite" force on the ship will partially cancel out the impulse it gained from the initial launch.
Likewise, the gravitational acceleration on the ship and the ball from any external field will be initially identical (ignoring the finite size of the "ship"), and any difference will only become significant as they separate.

 Posts: 1
 Joined: Wed May 29, 2019 3:14 pm UTC
Re: WhatIf 0085: "Rocket Golf"
I know I'm about 5 years late to the party, but I couldn't find rules against thread necromancy after a quick perusal and I happen to be particularly interested in this one.
First of all, if we wanted to count the weight of the acetylene, the easiest method is probably to calculate the volume. According to the paper, 400 cubic centimeters of acetylene were used to propel 50g to 140 m/s, and it so happens that the 400 cm^3 was reasonably close to or slightly higher than ambient pressure at 7000 feet, so 1 atm is a good guess for its pressure.
If the bag of golf balls has a diameter of 150 miles, according to CasualSax's post, there are 174,732,028,508,854,060,655 golf balls. However, Randal accounted for the fact that golf balls don't pack perfectly together, and CS did not, meaning that you could unpack the volume into 1.74x10^20 golf ballsized volumes, but you couldn't fit that many inside. Randall estimated that only about 65% of the total volume was actual golf ball, so only 65% of CS's number would fit inside, or about 1.136x10^20.
Assuming each to weigh 46 grams, that means we need to move about 5.22 x 10^21 grams of golf ball. Using a little unit analysis, 113.6 quintillion golf balls x 46 grams/golf ball x 400 cm^3/50 grams = 4.18x10^22 cm^3, or 4.18*10^16 cubic meters. That's a sphere about 215.3 km in radius, or 267ish miles in diameter (a lot wider than the bag of golf balls!) and Wolfram Alpha helpfully points out that it's about 3.1% of the total volume of the world's oceans. Under atmospheric conditions (70*F, 1 atm), acetylene has a density of about 1.084 kg/m^3, so our bag of acetylene would weigh 'only' about 4.53 x 10^16 kg, compared to the golf balls' 5.2 x 10^18 kg.
However, acetylene is rarely stored at atmospheric pressure; the acetylene used in the experiment was sourced from a welding tank of the sort available at any metalworking shop. Some quick web searches reveals that this sort of tank holds 10 cubic feet (.28 m^3) of pressurized acetylene at 200 atm. With some simple calculations, the 4.18*10^16 m^3 of the 1 atm bag would fit into a tank with a volume of a 'mere' 2.1x10^14 m^3.
A typical small welding tank weighs about 3.2 kg when empty. Assuming the weight of the empty tank scales with surface area (treating it as a hollow shell), it follows the square of the radius and not the cube like the volume does (if we keep the height/radius ration constant, we effectively make height a function of radius, turning the cylinder surface area equation into a quadratic in 'r' and the cylinder volume into a cubic). Increasing the volume by a factor of 7.5*10^14 (the simple ratio) increases the radius by a factor of about 90,850, so it increases the surface area and thus the weight by a factor of 8.25x10^9. The tank itself would thus weigh 2.6x10^10 kg, a negligible fraction of the gas's weight, which is only about a percent of the golf balls' weight.
Also, I am the E.D.S. Courtney who coauthored the paper maybe I should have mentioned that first, and am proud of being directly responsible for ensuring that the scientific name of a common potato was included in the paper's method  still find it funny. I can confirm that we are indeed a family of physicists who managed to turn screwing around with potato guns into a career, though most of the work has either focused on the 'gun' aspect and worked with firearms ballistics, or made a career out of building potato cannons sans potato and calling them shock tubes. I can't post links yet, but a quick google search for "Shock tube design for high intensity blast waves for laboratory testing of armor and combat materiel" will turn up a Defence Technology article I and my parents (she IS a Courtney by marriage) published a few years back; a quick inspection of Figure 1 will reveal that the shock tube designs used are all potato cannons with varying combustion chamber and barrel diameters.
We actually tried using the 'bottleneck' design in Figure B to launch vegetables; not potatoes, as a two inch barrel does not lend itself to airtight seals with potato projectiles, but chunks of other 'solid' vegetables such as turnips. We ended up with mostly mush, probably due to the inflexibility of the steel compared to the more typically used PVC; so if you're going to add a projectile to acetylene spud guns, make sure you use PVC and reinforce it (tightly wrapping Kevlar around the barrel would probably suffice, or fishing line if you don't have Kevlar) for maximum distance and minimal shards of plastic.
First of all, if we wanted to count the weight of the acetylene, the easiest method is probably to calculate the volume. According to the paper, 400 cubic centimeters of acetylene were used to propel 50g to 140 m/s, and it so happens that the 400 cm^3 was reasonably close to or slightly higher than ambient pressure at 7000 feet, so 1 atm is a good guess for its pressure.
If the bag of golf balls has a diameter of 150 miles, according to CasualSax's post, there are 174,732,028,508,854,060,655 golf balls. However, Randal accounted for the fact that golf balls don't pack perfectly together, and CS did not, meaning that you could unpack the volume into 1.74x10^20 golf ballsized volumes, but you couldn't fit that many inside. Randall estimated that only about 65% of the total volume was actual golf ball, so only 65% of CS's number would fit inside, or about 1.136x10^20.
Assuming each to weigh 46 grams, that means we need to move about 5.22 x 10^21 grams of golf ball. Using a little unit analysis, 113.6 quintillion golf balls x 46 grams/golf ball x 400 cm^3/50 grams = 4.18x10^22 cm^3, or 4.18*10^16 cubic meters. That's a sphere about 215.3 km in radius, or 267ish miles in diameter (a lot wider than the bag of golf balls!) and Wolfram Alpha helpfully points out that it's about 3.1% of the total volume of the world's oceans. Under atmospheric conditions (70*F, 1 atm), acetylene has a density of about 1.084 kg/m^3, so our bag of acetylene would weigh 'only' about 4.53 x 10^16 kg, compared to the golf balls' 5.2 x 10^18 kg.
However, acetylene is rarely stored at atmospheric pressure; the acetylene used in the experiment was sourced from a welding tank of the sort available at any metalworking shop. Some quick web searches reveals that this sort of tank holds 10 cubic feet (.28 m^3) of pressurized acetylene at 200 atm. With some simple calculations, the 4.18*10^16 m^3 of the 1 atm bag would fit into a tank with a volume of a 'mere' 2.1x10^14 m^3.
A typical small welding tank weighs about 3.2 kg when empty. Assuming the weight of the empty tank scales with surface area (treating it as a hollow shell), it follows the square of the radius and not the cube like the volume does (if we keep the height/radius ration constant, we effectively make height a function of radius, turning the cylinder surface area equation into a quadratic in 'r' and the cylinder volume into a cubic). Increasing the volume by a factor of 7.5*10^14 (the simple ratio) increases the radius by a factor of about 90,850, so it increases the surface area and thus the weight by a factor of 8.25x10^9. The tank itself would thus weigh 2.6x10^10 kg, a negligible fraction of the gas's weight, which is only about a percent of the golf balls' weight.
SCSimmons wrote:The Courtneys of the potato cannon research prove to be more interesting than expected on further research. When I saw that their last names were both Courtney, I questioned whether this was really a coincidence, and I'm pretty sure it's not. Michael Courtney has published papers, besides this one where he collaborated with E.D.S. Courtney, with Amy Courtney and Elya R. Courtney. Michael and Amy are listed as the principal researchers for BTG Research, a ballistics engineering company.
So, it sounds like there's a family of physicists out there who have managed to turn screwing around with potato guns collectively into a career. I've found my new personal heroes. Breakfast conversations must be fascinating, if sometimes frustrating. "What, hash browns again?" "What else am I supposed to do with the spent ammo?"
As an aside, Amy earned a B.S. in engineering mechanics from Michigan State one year before I got my physics B.S. from there. It's a big school, but it seems really unlikely we never had a class together; neither her name nor her face ring any bells for me, but her name probably wasn't Courtney back thenI expect she's a Courtney by marriage. Lucky girl.
Also, I am the E.D.S. Courtney who coauthored the paper maybe I should have mentioned that first, and am proud of being directly responsible for ensuring that the scientific name of a common potato was included in the paper's method  still find it funny. I can confirm that we are indeed a family of physicists who managed to turn screwing around with potato guns into a career, though most of the work has either focused on the 'gun' aspect and worked with firearms ballistics, or made a career out of building potato cannons sans potato and calling them shock tubes. I can't post links yet, but a quick google search for "Shock tube design for high intensity blast waves for laboratory testing of armor and combat materiel" will turn up a Defence Technology article I and my parents (she IS a Courtney by marriage) published a few years back; a quick inspection of Figure 1 will reveal that the shock tube designs used are all potato cannons with varying combustion chamber and barrel diameters.
We actually tried using the 'bottleneck' design in Figure B to launch vegetables; not potatoes, as a two inch barrel does not lend itself to airtight seals with potato projectiles, but chunks of other 'solid' vegetables such as turnips. We ended up with mostly mush, probably due to the inflexibility of the steel compared to the more typically used PVC; so if you're going to add a projectile to acetylene spud guns, make sure you use PVC and reinforce it (tightly wrapping Kevlar around the barrel would probably suffice, or fishing line if you don't have Kevlar) for maximum distance and minimal shards of plastic.
Who is online
Users browsing this forum: No registered users and 8 guests