A Question of Orbital Velocity

Before I take a shot at answering your question, can you clear up your premise for me? You wrote:

"If an object is travelling between 7 km/s and 11 km/s, perpendicular to the pull of gravity..."

Are you talking about an object orbiting the Earth, or the Sun, or something else?

"then as I understand, Newton says that this object will follow an elliptical path which is largely circular at this velocity. It is considered to be "in orbit" at this velocity"

Are you saying here that at this speed, an orbiting object will have an almost circular orbit, which is actually slightly elliptic?

"If a large disk were made to rotate, such that the average linear velocity of the disk were between 7km/s and 11 km/s, would it not be considered to be 'in orbit' as well?"

What do you mean by linear velocity of a disk here? Do you mean a disk spinning about its own axis or about some other axis?

Hi Roger,

1. Yes, orbiting earth, initially. These numbers of 7km/s and 11 km/s are the numbers that I was able to find representing escape velocities, first to orbit, and secondly, to leave the earth's gravity. (17,000 mph, and 25,000 mph)

2. Yes again, exactly as you say. (zero circular velocity, I believe it's called. the lowest possible orbit)

3. Linear velocity could also be called tip speed.. except the average might be found at approx 70% of the radius (rms?) If the disk were a wheel rolling on the ground, the tip speed would be found by taking the distance travelled per unit of time, and dividing by the circumference of the wheel. The average (of every particle in the wheel) would be something close to the rms value. (as I understand it)

Thanks,

Chris

Chris,

Thanks for the clarifications. I'm still a little confused. Is the disk spinning like a top? or is it rolling?

To get started, I may tell you some stuff you already know in order to get on the same page:

In physics, you can break the same equations in terms of Linear and Rotational Motions:

Linear Motion.....................Circular Motion..........................Conversion

x (Distance)......................θ (angle)..................................x=rθ (r is radius)

v (Linear Velocity)..............ω (Angular Velocity)...................v=rω

a (Linear Acceleration)........α (Angular Acceleration)..............a=rα

m (mass)..........................I (moment of inertia)..................I=mr² (for circle)

F=ma............................... F=Iα.......................................F=ma=mv²/r

KE=¹/₂mv² ............................KE=¹/₂Iω² ...................................KE=(¹/_²)mv²/r

The table above can help us calculate escape velocity and also lends a little insight into how linear and circular motion are related. Notice the prominent role "r" (the distance the rotating object is from the axis of rotation) plays in the conversion between Linear and Circular motion.

Escape velocity is determined by setting the Kinetic Energy (KE) of the object in orbit equal to the gravitational potential energy. This basically means the energy balances out and the object can "escape" the Earths gravity. So lets do that:

Notice the little m cancels out and only the mass you're trying to escape matters. Notice the "r", that's the distance the orbiting object is from the center of the Earth. The larger the distance, the smaller the escape velocity.

So when you say a disk rotating quickly, it matters what its axis is, since if it has a different rotational axis than the orbiting spacecraft in the example above, then it will have a different value of "r".

So please clarify how the object is rotating so I can understand what you're asking.

Roger,

My model is that the disk is spinning like a top, in free space, just above sea level (not computing any atmospheric resistance at this point). It's axis of rotation is along a radius from the center of gravity of the planet. The direction of motion of all of the particles of the rotating body are perpendicular to the pull of gravity.

Chris

Ok, that's what I thought, here's the problem:

In the escape velocity equation in the earlier comment, the velocity is calculated using r=radius of the Earth, for the object thats orbiting. That's because as this object orbits, it makes a big circle about the center of the Earth.

The top when it spins makes a circle about its own center, not the center of the Earth.

To calculate escape velocity, we used the radius of the Earth in our calculation because: radius of Earth x 2π = 1 orbit. Notice for a spinning disk: radius of Disk x 2π = 1 rotation of the disk.

Essentially you are saying the length of 1 rotation of the disk = the length of 1 orbit around the Earth. The point of course is not all rotations are the same, so we can't interchange them. It's not enough that an object is rotating at a certain speed, it also has to be in a particular arrangement with respect to the Earth. A spinning top is not in that arrangement.

Roger,

This is the point where I have run into trouble before, but mostly because I am stubborn... :)

Lets say that you are going for a drive in your space ship, so you launch yourself into orbit, and you are travelling along the path of the 'great circle'. Then you decide to take a left turn, so you apply a 'horizontal' thrust to the right, which applies a vector to the left, and now your 'great circle' is bent into a 'great cone', where the apex is still the center of gravity of the earth.

As you apply more and more thrust, your 'turning radius' becomes tighter and tighter, but you are still 'in orbit. You still must calculate the r to the earth but also the r of your turn. I don't see that one negates the other.

As you are driving along, and increasing your left turn, you encounter another spacecraft, who is also turning left, such that your turn circle matches his, and you are now turning around opposite sides of a circle, but both in orbit. You would be km apart, diametrically opposed, but having essentially the same velocity vectors.

So what if you now connected the two spaceships with a big central pair of spokes.

Both craft would still be moving at 'orbital speed', but in a circle. As nothing happened to zero your initial velocity, you must still be in orbit, yet your path is along the same lines as the disk or hoop we were discussing.

This is my logic, that tells me it should work... even if I can't do the math properly.

I hope I have explained this well enough. It would be easier if I could show you my diagrams.

Chris

Stubborn is fine, don't sweat it. Let's flip it around, rather than I try to convince you of the physics, how about you try to convince me. Here's where I'm getting hung up on your idea.

I visualize orbits and escape velocity best with cannon and projectiles. I imagine a cannon firing a cannonball with a certain initial velocity. When the initial velocity is small as compared to the escape velocity of the Earth, the cannonball does the following:

The cannon is on the left and the two blue lines are different trajectories of projectiles fired by the cannon. Notice on trajectory is longer than the other, this is because it had a fast inital velocity. Now lets see what happens when we get close to escape velocity.

Now the path of the projectile is still a parabola, but the distance is large enough that the earth starts to curve away from it. If you fire the projectile far enough, the Earth curves away at exactly the same rate that the projectile falls towards it, so it never hits the ground (and so is in orbit).

So what happens if the cannon fires a projectile with an initial velocity much faster than escape velocity? Well, the exact opposite of the first example, instead of the Earth seeming flat because it's so large, now the projectile path seems flat because it's so much larger. So it looks like this (Path E below):

So the escape velocity is the speed at which the cannonball has to be fired in order to go far enough so that the Earth curves away from it.

Now what I can't see, is how can rotating an object on the surface of the Earth really fast produce the same effect?

Roger,

Thank you for your diagrams. The last picture with the mountain is one I am familiar with. This is the diagram, and explanations I have seen and used before. It typically states, along with this, that 'initial velocity' is the only requirement for orbit, so long as it is perpendicular to the pull of gravity. My premise is based on this idea.

Secondly, I disagree with the notion that average velocity in a spinning disk cancels out. Firstly because there is no 'collision' between the particles. Secondly, if you put your fingers in an average fan, the average length of your fingers will be shorter. The disk is still in motion, and that motion is a velocity. If that velocity value is equal to orbital velocity, it will be in orbit.

As a better experiment, rather than dropping a disk, how about a large bubble chamber, where the heavy particles are injected horizontally with respect to gravity, and the magnets turn the particles into large circles (rather than spirals), and a 'horizontal view' is taken, rather than a top view, I think you would see a trajectory which would have the particles moving away from the center of earth's gravity, proportional to their velocity above orbital velocity.

Thank you for all your responses. I appreciate the supporting math.

As I said before, I am not a mathematician, but I would like to take your equations and write a vb program with them. In order for the equations to make sense, i would need units for the variables that I can make sense of. Can you add this to the list of variables and equations you provided?

Chris

I told you I'm not trying to convince you of anything. Those diagrams are how I think about escape velocity. If you truly want to explain to me how your idea works, explain it in a way I can understand. Relate it to the diagrams I put up. For instance:

You wrote:

"It typically states, along with this, that 'initial velocity' is the only requirement for orbit, so long as it is perpendicular to the pull of gravity."

Based on the diagrams I provided in my comment, I don't see why the initial velocity has to be perpendicular to the pull of gravity. It looks to me that the cannon could fire at any angle (except into the ground), and if the initial velocity is large enough, it would achieve orbit (or escape the Earths gravity).

Do you agree?

Roger,

Of course you are right.

I just focus on the horizontal because that is the way i see the particles moving in a disk.

I agree that the path is elliptical, and maybe as velocity is associated with a direction, maybe I should say that "Speed is the only requirement for orbit"

Regard 'the way that an object rotating on earth can produce the same effect',

I think the argument is the same as how lift can be generated from a rotating wing, versus a fixed wing. If an object with orbital speed can achieve orbit, so can a rotating object with orbital speed. I have used the word 'average velocity' perhaps incorrectly. To me, to say that 'the average velocity is zero' may be mathematically correct, but it still doesn't stop a helicopter rotor from producing lift. In a helicopter rotor, the primary variables for lift are essentially identical to that of a fixed wing. The closer to the root of the blade, the slower the velocity. The lift increases along the length of the blades, because they are moving faster. Any object that has orbital speed, regardless of direction, will be in orbit. Subsequently, I have long wondered if a spinning disk would hover.

So, just as you pointed out, direction doesn't matter. I agree with this. I think this is exactly the point why a spinning disk, above escape velocity, should move away from the source of gravity (acceleration field.)

Chris

You Wrote "I think the argument is the same as how lift can be generated from a rotating wing, versus a fixed wing. If an object with orbital speed can achieve orbit, so can a rotating object with orbital speed. I have used the word 'average velocity' perhaps incorrectly. To me, to say that 'the average velocity is zero' may be mathematically correct, but it still doesn't stop a helicopter rotor from producing lift. In a helicopter rotor, the primary variables for lift are essentially identical to that of a fixed wing. The closer to the root of the blade, the slower the velocity. The lift increases along the length of the blades, because they are moving faster. Any object that has orbital speed, regardless of direction, will be in orbit."

A rotating wing, such as used on a helicopter, creates lift by creating a large pressure difference between the air above the blades and the air below the blades. An object orbits because the cannon fired it so powerfully that it flew right off the Earth. I don't see how these two things are the same thing, they seem different.

Please explain how these are not different so I can understand how your idea would work. No math is necessary.

You wrote: "Based on the diagrams I provided in my comment, I don't see why the initial velocity has to be perpendicular to the pull of gravity. It looks to me that the cannon could fire at any angle (except into the ground), and if the initial velocity is large enough, it would achieve orbit (or escape the Earths gravity).

Do you agree?"

I am agreeing with you. Any angle will do. including the sum of angles when moving around a circle, such as a disk. In a disk, the particles which make up that disk, are in motion, and if they are moving above orbital speed, they will be flying off the earth.

Chris

You Wrote "I am agreeing with you."

That was two posts ago, my new question was different, please tell me how a Helicopter creating a pressure difference is the same as a cannon firing a projectile. If you explain this, I might understand your idea better. I'm trying to understand your idea. Please answer my new question.

Roger,

I only used the helicopter reference in comparison to a fixed wing. (not a cannon)

So then a helicopter has nothing to do with your idea? Wouldn't the spinning rotor of a helicopter produce the effect your talking about? Or does it have to be a spinning disk? All the logic you presented seems like it would apply to helicopters as well. Does it?

You wrote: "only the mass you're trying to escape matters"

I'm sorry, I don't understand. Did you mean to say more or is this an answer to my question?

Sorry Roger.

I don't think the shape of the object matters. If it rotates, the forces acting upon it will be the same, just varying in magnitude, based on angular velocity. It could apply to disks, rotors, rocks, or electrons. I view them all the same. Based on your previous posts, I was certain you knew this better than me. As I said earlier, I am excluding atmospheric conditions from my examinations, so it isn't based on any aerodynamic profile. The physics of this are the same on the moon, or free space, as on earth at sea level, if atmospheric conditions are excluded.

If I have upset you in some way, I apologize. it was not my intention.

Thank you,

Chris

Don't worry, I'm not upset at all.

My question now is, why wouldn't a helicopters just go into space. If any spinning object will have this effect, wouldn't helicopters have it too? Helicopter blades spin really fast.

Roger, I answered you.

"I don't think the shape of the object matters. If it rotates, the forces acting upon it will be the same, just varying in magnitude, based on angular velocity. It could apply to disks, rotors, rocks, or electrons." I think this includes helicopter rotors.

I know this sounds crazy, but I haven't been able to dismiss it. If I could find a way to calculate the forces, in a way that I can reasonably compare the centrifugal/centripetal forces to the material strengths, and find the design configurations where the extreme forces involved can be countered, i think it would be possible to test this hypothesis.

I think that as the radius of the Object increases, the angular velocity will decrease, while the tip speed remains constant, and consequently, the centrifugal forces will decrease. So increasing the radius is one way to counter the forces. I think that using ultra-strong composite fibre constructions is another way to increase strength.

Chris

I'm sorry, you're misunderstanding me.

You are saying that the helicopter would produce the force. So my question is, why don't helicopters go into space? If they are producing this force, there should be no altitude restriction for helicopters. So why is it helicopters have altitude restrictions. Why doesn't this force you are mentioning show up?

Helicopters have altitude limitations because they operate on aerodynamic principles, and as they rise, the air becomes thinner, therefore they run out of operating medium. Secondly, the rotors are not made strong enough to be able to operate at the rotor speeds that thinner air would require.

If you have a rotor which could withstand the centrifugal forces generated when the tip of the rotor moves at a circumferenctial speed greater than 7km/s, they would see the effect. However, if the circumference of a helicopter rotor is say 30 meters, then the rotor will have to turn 233.3 revolutions per second, which would generate considerable centrifugal forces.

If the helicopter rotor were 7 km in circumference, then it would have to turn once per second, and the centrifugal forces would be lower (in my opinion) and the effect would be evident.

Chris

You Wrote: "If you have a rotor which could withstand the centrifugal forces generated when the tip of the rotor moves at a circumferenctial speed greater than 7km/s, they would see the effect. However, if the circumference of a helicopter rotor is say 30 meters, then the rotor will have to turn 233.3 revolutions per second, which would generate considerable centrifugal forces."

So what you're saying is there wouldn't be any effect until you hit 7km/s. The antigravity effect would be sudden? From no effect to hovering?

Roger,

the effect would be identical to the way a space craft acheives orbit, and would be experiencing 'the effect' to the same degree a space craft does as it increases it's velocity towards escape velocity.

It is the same as if the orbital path of the space ship were curved into a circle horizontally.

another crazy idea for experimentation might be to create two matching circular particle colliders, where the particles are not vertically constrained, and place one on earth, and one on the moon. the test would measure the amount of vertical deflection, or actually lack of vertical deflection in the particles which are being turned into a circular path.

Chris

Quoting Chris: "It is the same as if the orbital path of the space ship were curved into a circle horizontally."

If you understand orbital mechanics, you will realize the futility of this exercise. A spaceship flying in 'horizontal circles' above a planet is one enormous waste of energy without any benefit!

"another crazy idea for experimentation might be to create two matching circular particle colliders, where the particles are not vertically constrained, and place one on earth, and one on the moon. the test would measure the amount of vertical deflection, or actually lack of vertical deflection in the particles which are being turned into a circular path."

Yep, this one is really crazy. Particles in circular accelerators are showing no vertical deflections above the 'straight, flat circle', even at 99.5% of c. They do not have to be 'constrained' vertically!

Jorrie

Okay, thanks Jorrie

Dear Mr. ROGER PINK/ Mr.CHRISG288

QUITE intresting subject appearing in this forum.

A bout 18 months back I raised one point ( which is my doubt) in this forum about space craft/satellite orbit.

Again it may look silly/irrelevent.

The space craft or satellite is relatively a TINY object orbitting in terms of size and speed when compared to the Planets in the solar system. We all have studied NEWTON's LAW. The Newton's First Law states that "EVERY BODY CONTINUES TO BE IN ITS STATE OF MOTOIN or REST, and UNLESS compelled by an external force there is NO change in the Motion/Direction of movement or Rest.

If such be the case, my doubt is - the satellite/space craft should coninue in its straight line and not circular orbit or elliptical orbit.

Which force is keeping the elliptical/circular orbit.

There was only 3 responses and I will try to retrieve from my archives and post here.

DHAYANANDHAN.S

"If such be the case, my doubt is - the satellite/space craft should coninue in its straight line and not circular orbit or elliptical orbit."

G'day dhayanandhan

There's a simple one word answer to this and the word is GRAVITY. As the spacecraft travels along in orbit the Earth's gravity keeps trying to pull it back towards the ground, however, as the craft moves in relation to the surface of the earth the direction of that gravitational force changes as it always pulls (actually space pushes it, but that relativity and were talking Newtonian physics here) towards the centre of the Earth's mass. Since the earth is pretty close to being spherical if you add the ever changing direction of gravity to the motion of the satellite you get the equation for either a circle or ellipse depending on your starting point an initial motion vector.

I'm sorry I can't give you the mathematical proof as my mathematics is just a tad rusty after 30 years, but Wikipedia has a reasonable article on the subject of Orbital Mechanics, just skip over the bit about relativity because it will likely confuse the hell out of you like it does most people including myself most of the time. If you want to understand yow relativity comes into it then I recommend asking Jorrie he's actually written a eBook on the subject titled "Relativity for Engineers" which I thoroughly recommend.

By the way, there are NO STRAIGHT LNES IN NATURE ANYWHERE, not even the horizon or the path a beam of LAZER light travels along everything is a curve of some description, a straight line is something we have made up to help explain things mathematically it doesn't actually exist.

I hope I have helped explain things and not confused things further, but we are talking about a chapter in the very large subject of rocket science here and that's about as complicated as things get.

Dear Mr.masu,

Thank you very much for your posting. My doubt is cleared. I will down-load and read the details of the references referred by you and study.

Any doubt I will put a Mail through CR4 Personal Messaging and indeed, I am thankful to you.

Thanks,

DHAYANANDHAN.S

Chris wrote: "Both craft would still be moving at 'orbital speed', but in a circle. As nothing happened to zero your initial velocity, you must still be in orbit, yet your path is along the same lines as the disk or hoop we were discussing."

The orbital scenario you sketched is not possible to achieve in the way you said. Before the two craft could be turning in circles around a hypothetical center point, they would start to move in differently orientated orbits around Earth. This is straight orbital dynamics. So you cannot link them up with a piece of metal.

What you can do it put the two craft in identical orbits, but at opposite sides of Earth, both circling Earth's center of mass. You can even construct a hoop that spans around Earth at orbital height and rotate it at circumferential speed equal to orbital velocity. But what is the point?

Do you try to create something like anti-gravity that way? Forget it!

Hello Roger

Re escape velocity

For the case of escape velocity from the Earth's surface (which is what we're usually interested in) it's simpler to express it in terms of g, v_e = √(2gr) by eliminating M and G using mg = mMG/r², or calculating energy change from F = mg(r/R)², where R = radius > r. Not everybody knows M and G offhand, but most scientists and engineers know g (or ought to!)

Above formula only works when r = surface radius, (whereas yours works for any r >= surface radius) but can readily be generalised to any radius R, giving v_e = r √(2g/R) (where r still = surface radius)

If I understand you correctly you were asking why a disk that was spinning around a stationary vertical axis that with average linear speed greater than 7 Kms^-1 isn't in orbit.

I think what you are doing is confusing VECTORS and SCALARS. If you sum up the velocity vectors for each point on the rotating disk you get zero. Remember that a vector is a speed with a direction. For every point on the disk there is and opposing point on the other side of the disk going in the opposite direction, hence the velocity is cancelled out. Since the average velocity of the disk is zero then it isn't in orbit.

What you were doing was calling the average speed of the disk a vector when it is actually a scalar.

Chris

The previous responders are correct. It won't work. However, there is a very easy way to test your concept. Take a solid steel cylindrical disk with a solid central shaft made as symmetrical and balanced as you can. It should only weigh a couple of pounds (a few kilonewtons). Spin it to a few hundred rpm, then let it drop from a height of about 100 feet as it is spinning, measuring the time it takes to fall from release to impact. Repeat this many times and find the average drop time. Then choose a different spin rate, perhaps 3000 rpm and repeat the experiment. If you like, you can vary the spin rates, drop heights and object sizes to get a more complete experimental data set. What you will find is that other than experimental error, there is no significant difference in the fall time due to the spin rate.

From an experimental standpoint, the maximum spin rate is determined by the strength of the material of the spinning disk, so anything beyond that maximum is untestable and therefore experimentally unknowable. From a mathematical point of view, the previous responders are correct based upon mathematical representations of observable physical phenomena such as planetary orbits, spacecraft behavior, spinning gyroscopes, etc.

By the way, what you are describing is essentially a large gyroscope and gyroscopes just don't behave in the manner you describe.

Interestingly, this is something I've personally been thinking about for about 10 years now. I wish I ran into this thread 4 years ago!

Well I did want to comment on a couple of things here.

Firstly, I noticed JohnB's comment because I thought the same thing a few years ago. I figured that if a ring spinning rapidly enough could in essence levitate, then I would have guessed at that time that the ring would experience an increase in the time it took to fall to the ground as it rotational speed increased. Going along with that, I also thought that you would see a decrease in weight (not mass, but weight). However, after giving it some thought, I realized that this would not occur at all.

Here's why:

Using the illustration of the cannonball, we know that the ball is always falling towards earth with the same acceleration. As long as the ball does not reach orbital speed, it will always end up hitting the surface of the earth in the same amount of time. Once it reaches orbital speed, we know that it will never TOUCH the surface of the earth, but the ball is still falling towards the earth at the same rate. It's just that it's moving forward so rapidly that it covers enough distance in the time it would otherwise impact that surface of the earth that it ends up never touching the surface.

With that said, regardless of whether or not a rapidly rotating ring could essentially levitate, it seems clear to me that it would not show any change in weight or the time it takes to impact the surface of the earth until orbit velocity is reached. At this point (assuming that it would achieve "orbit"), I would expect the ring to suddenly become weightless and not fall towards the earth at all until its rotational speed decreases below orbital velocity. In essence, what I suggest here is that NO change in weight or rate of fall would be observed at all until the instant the ring reaches the necessary speed of rotation for "orbit".

With respect to materials strong enough to test this idea, you would need materials much stronger than carbon fiber. Also, the size of the disk itself would not appear to matter, since the stress on the ring caused by it's rotational speed would only change with the distance the ring is from the center of mass of the planet. That is to say, I can make the radius of the ring as large as I want, but the stress on the ring due to its rotational speed would only vary based upon altitude.

I spent 15 years thinking about it... and could not find anybody that could explain to me, in a way that made sense, why it wouldn't work.. When I did finally recieve that information, it was rather humbling.

... because the the man told me in one sentence, upon first hearing, why it wouldn't work.

He said "regardless of how fast it is spinning, the center of the disk does not coincide with the center gravity of the planet, and therefore will fall. In your comparative analysis of the rocketship, the center of rotation does coincide with the center of gravity of the planet."

I was blown away... all those years, I had focused on the notion that 'velocity alone was sufficient...'

good to know. I wouldn't be who I am without that little obsession...lol

Chris

Hey Chris, glad to see that you've gotten some more info since then. It seems logical to conclude that since the center of mass of the rotating ring is not moving with respect to the center of mass of the earth, then it would not levitate.

Interestingly, a spacecraft actually can use thrust to create what I like to call a displaced orbit, but officially this is known as a non-keplerian orbit. There isn't a huge amount of information on the web about this as far as I have found, but non-keplerian orbits are definitely a topic of serious research. Your analogy of a spacecraft using thrust to make it turn as it orbits is actually exactly how a non-keplerian orbit would work. Essentially, it's a "great cone" orbit.

Chrisg288 getting back to your original question. Lets neglect the effects of atmosphere, rotation of the earth etc and assume a simple planet with no atmosphere.

I believe what you are describing is something like this. There are two identical objects tied together with a rope and are rotating around the center of the rope at 7 Kms^-1 and the question you are asking is that since that are traveling at orbital velocity then why aren't they in orbit?

Well as I said earlier it a question of VECTORS and SCALARS. While the two objects are connected to each other you need to add the two instantaneous velocity vectors to get the true velocity vector for the system. Since the object on the other end of the rope is traveling in the opposite direction at the same speed they cancel each other out.

If the rope were to break however then the two object would no longer interact and you would no longer need to add the two vectors. In this case given the conditions above both objects would end up in orbit, albeit in opposite directions.

In other words while they are mechanically connected then you need to add the velocity vectors and that is the reason your disk rotating with an average rotational speed greater than 7 Kms^-1 doesn't end up levitating.

Does that clear up your question?.

Good explanation, masu! It reminds about that crazy idea of the 'space earovator', a form of space elevator. The huge aerofoil rotates around one end and aerodynamic lift pushes its other end into space, with its tip moving at near 8 km/s. Objects are launched into orbit by making them 'slide' up this huge rotating 'aerofoil'. See this Wikipedia article.

Chris, note that this contraption makes use of aerodynamics for lift and not your idea.

Regards, Jorrie