Bowling Ball Straight What kind of bowling ball is good for me?
He! I am searching for a new bowling ball, can you help me find one that is right for me? I am an inside shot bowler. So basically I start out in the middle and my arm is over the last arrow on the right.Sometimes I will throw the ball at an angle or straight depending on the lane. So what ball is good for me?
Any help is appreciated, thank you!
You should try out a Columbia 300-Freeze or a Raw Hammer-Sting I think it's called
Drilling A Bowling Ball What does drilling a bowling ball mean?
So I want to get a bowling ball but I'm kinda confused with all the different kinds and everything that has to be done to it. I usually use the balls at the bowling alley but I used my friends ball for a week and I got a higher avg by hooking it. I look online and there's one ball that I want to get but I'm a little cnfused about the drilling. Doesn't that just mean you just drill it so you have fingerholes? Then why is there a separate option asking if you want it drilled. And if I want to add finger inserts to it, would I have to go to a local shop to get that done?
Sorry if it's kinda a stupid question.
Your local bowling store can drill the holes for your fingers. This is done custom, to fit your hand.
Mom and I gave Dad a new ball and bag for Xmas, and we went to the store at the bowling alley.
We bought an AMF, but Brunswick is a good brand, too.
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I would like to know how could own my own pro shop in a bowling alley to drill balls and how i could be taught
I love bowling and i honestly never considered owning my own pro shop and would like to know how i could get taught to drill bowling balls and how much do you think could be made drilling bowling balls and selling other accessories at a pro-shop in a bowling alley
It really depends on how old you are. If you are young you can still get into it. If your older it will be very difficult. Most of the pro shop opperators are pros or ex pros that have had many years experience with the technical side of the equipment. It will take YEARS to learn to drill correctly. you will also have to know everything about the balls and what to reccomend for people.The pro shop that I help at brings in 6 figures a year. Its good money but not something your going to get rich off of. It also takes a lot of time and dedication.
Bowling Ball Release How would I calculate the time it takes for a 12 lb bowling ball to sink to a 5 mile depth in the ocean?
If I was on a boat at a location in the ocean that was 5 miles deep, and I released 12 lb bowling ball from the surface, how do I calculate the time it takes for it to reach the bottom?
I'm guessing that the variables include the viscosity of the water and the constant of gravity and the surface of the ball itself.
And before anyone comments on it , this actually isn't a homework question, despite the fact that it sounds like one. This question is simply out of personal curiosity.
Well, an object's buoyancy is equal to the weight of the fluid it displaces, and acts with a force equal to that weight. So, a bowling ball will sink with normal gravitational acceleration (9.8 m/s^2) minus the force pushing it up by the water it is displacing, until it reaches its terminal velocity, when it will to continue to sink at that rate until it hits the bottom. Kind of.
The density of sea water is highly variable, and taking pressure, temperature, and salinity into account makes a calculation like this very tedious. The best way to go about it is to assume an average density for seawater for the whole trip down. The calculation will be simpler, but the answer should be moderately accurate. Water at the surface has a density of about 1.025 g/mL, and deep, cold water about 1.05 g/mL, so 1.0375 g/mL is an acceptable value to use.
The best way to see how an object sinks is to calculate its net force. This is just the weight of the object (downward force) minus the buoyancy of the object:
Fnet= mg - pVg,
where m is the object's mass (5.44 kg for a 12 lb ball), p is the fluid density (1037.5 kg/m/s^s), V is the object's volume (4/3pi times it's radius [.109m for a standard ball] cubed), and g is the coefficient of gravitational acceleration (9.8m/s^2 on Earth).
So, Fnet = (5.44kg)(9.8m/s^2) - (1037.5kg/m^3)(4/3pi[.109]^3)(9.8m/s^2)
The answer, in a unit called Newtons ([kg m]/s^2), is 47.6839. But what does it mean?
The net force on the ball is 47.6839 Newtons toward the bottom (force is a vector, and therefore has directionality, but more simply, we know the bowling ball is sinking). Recalling another of Newton's equations will allow us to calculate the acceleration of the ball. Specifically, F=ma.
We already know the Force and the mass, so we can plug those in and solve for acceleration. To make things simpler, we'll move the formula around before we put the numbers in it.
F=ma, divide both sides by m,
and F/m= a. So if we divide the Force by the mass, we'll get the acceleration. This makes sense, remember our units of Force are newtons, kg m/s^2, so if you remove the kg, you just have m/s^2, the basic unit of acceleration.
So, 47.6839N/5.44kg = 8.765m/s^2 This means that every second, the ball will be moving 8.765 m/s faster than it was the second before.
Now, all we have to do is find out how long the ball will accelerate before reaching terminal velocity, and just put everything together.
The formula for calculating terminal velocity is
Vt= (sqrt[2ma/pAD])
where m is the objects mass, a is the object's acceleration, p is the fluid density, A is the surface area of the object (4pi[radius^2], and D is the drag coefficient (.47 for a sphere).
This means that the fastest the ball can sink is .378 meters per second, which seems about right. That's about a foot and 2.5 inches a second.
This low terminal velocity is convenient, since we know the bowling ball (because of its 8.765m/s^2) will reach it well in under a second, we can just assume this speed for the entire trip.
Now, all we have to do is see how long the ball takes to travel 5 miles (8,046.72 meters) at this speed.
So, a 12 pound (5.44 kg) bowling ball, dropped into 5 miles of ocean (8,046.72 meters) will accelerate until it reaches a velocity .378m/s, at sink straight to the bottom in 5 hours, 54 minutes, and 36 seconds.
