How does a directional drilling rig work. How does it drill horizontally out of a vertical hole.?

After spending a number of years as a petroleum geologist, I can give you the short answer. For the long answer, you will need a petroleum engineer.

Basically, an oil (or gas) drilling rig will drill a verticle hole for a specified distance. It uses a standard drilling arrangement, usually a tri-cone drilling bit on the end a hollow drill string. The "kelly table" on the floor of the drilling rig rotates the drill string and bit and the rig pumps drilling mud down the center of the drill string, through the bit (to cool it and transport rock debris out) and the mud flows back up to the drilling rig between the drill hole wall and the drill pipe.

Now comes the interesting part. When the verticle hole is done, they pull the drill string out of the hole and put a stearable sub and mud motor between the end of the drill pipe and the drill bit. The stearable sub allows the driller to point the drill bit at a slight angle from verticle. The mud motor (hard to describe) actually provides rotation to the drill bit powered by the force of the drill mud being pumped through it.

The driller will run the drill string back in the hole, points the bit at the correct angle and begins pumping drilling mud down the drill string. Now, the drill string cannot be rotated anymore, it must remain still, except for downward pressure. The mud motor rotates the bit and they begin drilling again. Drill pipe will bend, but not very much or you will get stuck. So the angle of deflection has to be small and over a long distance. But eventually the drill hole will be horizonal.

This is a very simplistic explaination of an incrediblly complex operation. The engineers, geologists, and tool pusher know exactly where they are going and how to get there.

BTW - Did you know that most deep oil and gas drill holes are not strtaight. They tend to corkscrew in the direction of drilling.

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How does a directional drilling rig work. How does it drill horizontally out of a vertical hole.?

How Does Directional Drilling Work

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Actually, that is incorrect. The initial angular velocity of the two balls is equal to the rotation of the earth so they will be moving together. Saying otherwise is like arguing that when you jump up and down, you will move really far to the side because the earth is rotating around the sun at about 30 km/s (not per hour, per SECOND). As far as the balls are concerned, the earth is a closed stationary system. Relativity would make the same argument. You mention the actual time it takes for this oscillation. I wonder if this is suppose to mean something or is just a red herring. To be perfectly honest, I have to disagree that no change in their distance will occur. When the two balls are at the center of the earth, they will probably collide. Of course, I am using the term collide very very loosely here. It is more like they keep getting closer until they are just barely touching, and then they are touching, and then they can't get any closer. So they spend some time stuck together before they reach the center of the earth. at this point, they can be treated as just one object because there is no tangental components to their velocities. They negate each other. If this had occurred in a fast collision where they separated immediately, then, you could argue they will have the same distance apart on the other side. However, the balls are too close together for that to happen. When the balls pass the center, their velocities are purely radial away from the center. They experience no forces to pull them apart. on the contrary, they are experiencing a force (gravity from the center of the earth) that pulls them together as well as back down. So, they should probably exit together with no distance between them. All of that was just theorizing and guessing though. Something like this is already way over simplified and impossible to test. I think my line of reasoning makes sense though. EDIT: Wow, there are a lot of spam posts here. Firstly, drilling through the Earth does not create a volcano. Those are formed at the meeting of two tectonic plates not at the north pole. These answers are really ridiculous. Secondly, it doesn't matter precisely how big the particles are. If they are they are the size atoms, peas, bowling balls, or even the moon, they will still collide. In an ideal system they must travel right next to each other the entire time pulling directly towards the center. They both will pass through the precise center at the exact same moment in time. However, assuming that their size tends towards zero, then their collision would be instantaneous and thus cause them separate out after they pass through the center. This way conservation of momentum will be conserved. Anyway, the falling down process becomes identical to flying up process. They will be the exact same distance apart by the time they come out. Actually, I want to try some calculations here. The arc length of the two balls starts out as 1 cm. Let's see, this equation states: ArcLength=Radius*angle where angle is constant throughout the dropping because all forces are radial, not tangental. So, the two particles would collide when the ArcLength is the width of one atom. I will be using the polar radius because the actual position of the tunnel was stated. angle = 0.01/Radius of Earth = 0.01/ 6,356,752 = 1/635,675,200 diameter of H is 50 pm = 50 x 10^-12 5x10^-11 = R / 635,675,200 or R = 5x10^-11 * 635,675,200 = 0.0317 m So, if we drop 2 hydrogen atoms at 1 cm apart and assume no interaction with each other as they fall, they will collide 3 cm before reaching the center of the Earth (or 30 cm is the atom was as big as cesium). Alright, so the question is whether or not that is a significant amount of time. To judge that we need to know how fast they are moving at that time. I just attempted to integrate the potential energy (f will mean integral from now on). Also, Me is the total mass of earth while me is the partial mass of some radius, x, of the earth. Me = k * Re^3 k = Me / Re^3 me = k * x^3 me = (Me / Re^3) * x^3 U= -f F dx = -f (G*me*m/x^2) dx = -G*m f [ (Me/Re^3) * x^3 /x^2] = -G*m * (Me / Re^3) * f x dx = " " * (1/2) x^2 | x=Re -> x=0 = " " * (1/2) * [ 0^2 - Re^2] = " " * (1/2) * (-Re^2) = (1/2)*m*G*Me/Re Conservation of energy: potential Energy, U, becomes Kinetic Energy, K. U = K = (1/2)m*v^2 = (1/2)*m*G*Me/Re v^2 = G*Me/Re (which are all constants) v = 7,918 m/s t = d/v = 0.01/7918 = 1.26 * 10^-6 seconds So we have to double the distance and the time because we are talking about falling to the center and then moving past. The question is do we think 2.5 x 10^-5 seconds and/or 6 centimeters are a significant time for two particles to be colliding when they are on the size of a Hydrogen atom or d = 50x10^-12 m. Well, this is a judgement call that you will have to make. If we were to take in real physics, the two particles would probably try react together or repel each other but we are ignoring that fact. Collision cannot be ignored in this case. there is no way you can say they don't collide. However, we don't know if they stay stuck together or not. IN CONCLUSION, my opinion is that no matter the size of the particle, they will collide at the center of the Earth and stay stuck together. However, there is ONE other answer. The two could collide and simply bounce back which means that their exit distance is going to be identical to the entering distance or 1 cm. These are the ONLY TWO CHOICES. I assume the correct answer would be the latter for any test or exam since they are likely to neglect how long the two collided particles are touching. I hope this helped.

I've used pnuematic piercing tools to drill small tunnels under roadways, although Ditch Witch (Charles Machine Works) makes a rig that looks like something out of Star Wars. You sit in the rig and it drills with a location beacon on the drill head you can steer the drill head as you sit in the rig staring at an LCD screen.

Directional drilling used special drills or water jets to punch holes horizontially ...under road ways, walks, etc., to set pipes, conduits, etc.