If so, where does the forward thrust vector come from?
On a sailboat it makes sense to me, because the sails are oriented at an angle to both the apparent wind and the direction of travel of the vessel, so the wind has a forward thrust vector as well as a sideways vector. But on a bike, the wheels are oriented parallel to the direction of travel. So where is the “forward thrust” coming from?
From the low pressure side of the wheel, just like on a sail from a sail boat.
On a sailboat the curve is built into the sail so that the deepest part of the sail is about 1/3 of the way from the leading edge. This brings the low pressure point near the front of the sail.
On a wheel, that cannot be built into it so the low pressure point is further back than on a sail, but if that low pressure point is lower than the high point on the other side and forward of the center of effort, then the low pressure will pull the wheel in that direction.
It will not be completely forward but the wheel will not slide sideways much like a sailboat resists going sideways from the sail effect of the keel under the boat.
If so, where does the forward thrust vector come from?
On a sailboat it makes sense to me, because the sails are oriented at an angle to both the apparent wind and the direction of travel of the vessel, so the wind has a forward thrust vector as well as a sideways vector. But on a bike, the wheels are oriented parallel to the direction of travel. So where is the “forward thrust” coming from?
Imagine that the disc is a wing (it’s not an airfoil, but even a flat disc can approximate a wing–heck, with enough power and the right pitch attitude, you can get an airplane to fly with wings that have no airfoil at all).
The wing (the disc) will develop lift, not perpendicular to the cross-section of the wing (the disc), but instead perpendicular to the angle of the relative wind.
Now imagine a situation when you are riding with a bike with a disc with the relative wind well off to one side or the other (in the real world, that will usually be when you are riding forward with a good cross-wind). Draw a diagram with some approximate force vectors and then it becomes easier to see how a component of the lift on the disc can actually propel you forward.
From the low pressure side of the wheel, just like on a sail from a sail boat.
On a sailboat the curve is built into the sail so that the deepest part of the sail is about 1/3 of the way from the leading edge. This brings the low pressure point near the front of the sail.
On a wheel, that cannot be built into it so the low pressure point is further back than on a sail, but if that low pressure point is lower than the high point on the other side and forward of the center of effort, then the low pressure will pull the wheel in that direction.
It will not be completely forward but the wheel will not slide sideways much like a sailboat resists going sideways from the sail effect of the keel under the boat.
If the wheel is stationary, and you apply a quartering headwind to it, the wheel will try to move forward then? I just don’t see any way that what you say can actually be true, since the angle of attack is zero. planes fly and sailboats go forward because there is an angle of attack for the wing / sail. The boat doesn’t go forward if the boom is pulled over the centerline of the hull too far.
If the wheel is stationary, and you apply a quartering headwind to it, the wheel will try to move forward then? I just don’t see any way that what you say can actually be true, since the angle of attack is zero. planes fly and sailboats go forward because there is an angle of attack for the wing / sail. The boat doesn’t go forward if the boom is pulled over the centerline of the hull too far.
The angle of attack of a disc is not zero when the relative wind is from anywhere other than from straight ahead.
angle of attack in relation to the direction of travel. that’s probably the wrong term for it.
I just don’t get it. I’ll have to go back and think about this some more, but I don’t get the “lift” component vector being perpendicular to the apparent wind. Why is that true?
I found this to be a helpful visualization:
.
angle of attack in relation to the direction of travel. that’s probably the wrong term for it.
I just don’t get it. I’ll have to go back and think about this some more, but I don’t get the “lift” component vector being perpendicular to the apparent wind. Why is that true?
The direction of travel is important to you, but it is not important to the wind, to the disc, or to the resultant forces.
Ultimately, the ‘lift’ component being perpendicular to the relative/apparent wind is not set in stone, but it is just a labeling convention that makes understanding what is going on a lot easier. For example, typically for an aircraft in level flight, there are 4 labeled forces (thrust, drag, lift, gravity) with the lift and gravity forces being both perpendicular to the relative wind and to the ground (in level flight). This simplified labeling convention can then be applied to a bike going forward at a constant speed with a steady crosswind.
Then, there is always this diagram (via the Flo website) that shows the vector of the sail effect (the ‘Y force’ vector drawn near the disc wheel) nicely:
my brain is hurting now, but what I’m confused about is where that “Y” component of the side force is coming from.
on that diagram, I think the “X” forces of the drag component must be the same thing. but why is the side force (lift) perpendicular to the apparent wind? I’m assuming that this holds for angles less than the “stall” angle.
the direction of travel matters because there needs to be a force component in that direction for the sail effect to work.
(as an aside, I’m talking about a “sail effect” in the sense that the disc will actually result in forward motion, not merely drag reduction, as that’s how I think a lot of people think of the term.)
Question then…would the most optimal way to take advantage of this be to just have your “gansta lean” going down the road for a tough cross wind…or would it be more optimal to “tack” the bike back and forth across the road.
Like, slowly drift in the direction of the effect. Then, if the wind lets up for even just two seconds move the bike in the opposite direction anticipating the wind to blow you again.
Obviously only for a closed course and nobody else around you.
my brain is hurting now, but what I’m confused about is where that “Y” component of the side force is coming from.
on that diagram, I think the “X” forces of the drag component must be the same thing. but why is the side force (lift) perpendicular to the apparent wind? I’m assuming that this holds for angles less than the “stall” angle.
the direction of travel matters because there needs to be a force component in that direction for the sail effect to work.
(as an aside, I’m talking about a “sail effect” in the sense that the disc will actually result in forward motion, not merely drag reduction, as that’s how I think a lot of people think of the term.)
Looking more closely at the Flo diagram, I think they mislabeled a few things.
I don’t think the sail effect claims there will be forward motion of a whole bicycle at high speeds (that would be a lot of drag to overcome, only that in certain conditions you can get a situation that gives negative drag (so a slight slight forward ‘push’) from a disc wheel.
I’m talking about wheel only, not the entire bike. Does it make sense that the wheel would move forward when presented with a headwind at an angle?
Does it make sense that a sailboat can go upwind when presented with wind at an angle? It does, because it happens.
Given the non-optimal shape of a disc wheel, it can’t happen in as many circumstances as on a decent sailboat, but, yes, negative drag, means that a wheel can indeed move forward.
yes it does, because I can visualize the force component vectors. the sail is at an angle to the direction of travel of the vessel, so it’s clear to me where the forward vector is coming from.
I can’t visualize a forward force vector for a disc wheel which is, by definition, parallel to the direction of travel.
***negative drag, means that a wheel can indeed move forward. ***
Got a demonstration of that effect?
Got a demonstration of that effect?
Me, personally ?
For one, I don’t own a disc.
But this effect has been repeatedly shown in wind tunnels.
yes it does, because I can visualize the force component vectors. the sail is at an angle to the direction of travel of the vessel, so it’s clear to me where the forward vector is coming from.
I can’t visualize a forward force vector for a disc wheel which is, by definition, parallel to the direction of travel.
You seem to be visualizing it as though the wind is coming from straight ahead, though, when that is rarely the case. If the wind is coming from either side, then the quartering effect on the plane of the disc is really not that different than the sail on a boat being at an angle to the direction of travel. There’s a little difference of course in that the sail can pivot so that it’s not in line with the center axis of the boat whereas the wheel of course is locked in straight, but that only means you can’t optimize the sail effect by changing the angle it meets the wind; the effect is still there at some level, however. And while the lift it creates may not quite be straight ahead in the direction of travel, it’s still in a forward quarter so that’s a negative drag.
yes it does, because I can visualize the force component vectors. the sail is at an angle to the direction of travel of the vessel, so it’s clear to me where the forward vector is coming from.
I can’t visualize a forward force vector for a disc wheel which is, by definition, parallel to the direction of travel.
You seem to be visualizing it as though the wind is coming from straight ahead, though, when that is rarely the case. If the wind is coming from either side, then the quartering effect on the plane of the disc is really not that different than the sail on a boat being at an angle to the direction of travel. There’s a little difference of course in that the sail can pivot so that it’s not in line with the center axis of the boat whereas the wheel of course is locked in straight, but that only means you can’t optimize the sail effect by changing the angle it meets the wind; the effect is still there at some level, however. And while the lift it creates may not quite be straight ahead in the direction of travel, it’s still in a forward quarter so that’s a negative drag.
no, I’m not. I used to sail when I was a kid, so I have some rudimentary understanding of how a boat actually works. A sailboat doesn’t do anything when the wind is coming from straight ahead or a few degrees to either side (roughly 15* or so, if memory serves), only if it’s coming from the side or stern.
I’m trying to visualize all of this using a quartering headwind (apparent angle).
yes it does, because I can visualize the force component vectors. the sail is at an angle to the direction of travel of the vessel, so it’s clear to me where the forward vector is coming from.
I can’t visualize a forward force vector for a disc wheel which is, by definition, parallel to the direction of travel.
You seem to be visualizing it as though the wind is coming from straight ahead, though, when that is rarely the case. If the wind is coming from either side, then the quartering effect on the plane of the disc is really not that different than the sail on a boat being at an angle to the direction of travel. There’s a little difference of course in that the sail can pivot so that it’s not in line with the center axis of the boat whereas the wheel of course is locked in straight, but that only means you can’t optimize the sail effect by changing the angle it meets the wind; the effect is still there at some level, however. And while the lift it creates may not quite be straight ahead in the direction of travel, it’s still in a forward quarter so that’s a negative drag.
no, I’m not. I used to sail when I was a kid, so I have some rudimentary understanding of how a boat actually works. A sailboat doesn’t do anything when the wind is coming from straight ahead or a few degrees to either side (roughly 15* or so, if memory serves), only if it’s coming from the side or stern.
I’m trying to visualize all of this using a quartering headwind (apparent angle).
Yes - it’s a quartering headwind; clearly you can’t sail directly into a headwind.
(well, maybe those sick America’s Cup boats can)
Depending on the boat & sail, you can probably pinch it to within 45 degrees on either side of the direction of the headwind.
Example - if the wind was from Due North, you could sail at closest to that probably being NW and NE.
The sail acts as an airfoil in this scenario, and the “lift” pulls you “forward”.
Now, how this does or does not relate to a disc wheel “sailing”, I cannot tell you.
The figure in post 7 is really nice, but what is missing, which could be causing some trouble, is the frictional force between the tire and the road that keeps the bike from just sliding in the sway direction in response to the lift from the disc wheel. In the figure they have decomposed the aerodynamic force from the wheel into aerofoil lift and drag components which are perpendicular and parallel to the apparent wind, as is standard.
You could also decompose that force into the reference frame of the bike, with a component in the surge (forward) direction, and a component in the sway (cross) direction. If you draw the balance of forces, the sway direction would need to be offset by friction, and the surge direction would eventually be balanced by aerodynamic drag once you have increased to a steady state speed.
In the case of a sailboat, it is really important to remember that the aerodynamic lift and drag from the sails is coupled with the hydrodynamic lift and drag from the hull, centerboard, and rudder. If the hydrodynamic forces are not present, the boat will just slide sideways. It is the coupling of the aero and hydro forces that allows sailboats to sail upwind, often at apparent wind angles much less than 45 degrees.
I have some slides on sailboat forces here: https://docs.google.com/...nIk/edit?usp=sharing
ETA: slides from a community education talk, so not very technical. Happy to try and explain more. I love this stuff!
Question then…would the most optimal way to take advantage of this be to just have your “gansta lean” going down the road for a tough cross wind…or would it be more optimal to “tack” the bike back and forth across the road.
Like, slowly drift in the direction of the effect. Then, if the wind lets up for even just two seconds move the bike in the opposite direction anticipating the wind to blow you again.
Obviously only for a closed course and nobody else around you.
For a wheel rolling down the road without a bicycle or rider, tacking is potentially a great strategy.