# Spin vs Speed

#### BogeyNoMore

##### * Ace No More *
Diamond level trusted reviewer
Hoping the physics heads out there can help me isolate how SPIN and SPEED (air speed, not a disc's "speed" rating) affect disc flight after the disc is released. Given: stability, speed, glide, how beat up it is, how you throw... all of it can have a major effect on disc flight. There's plenty of discussion on that here and elsewhere. Most of us know how a little (or a lot) of hyzer/ani affects our throws. (If you don't, I suggest you find out for yourself over your next few practice rounds, or better yet, in an open field).

For the purpose of this thread, please assume a disc is thrown perfectly flat, with neutral pitch (no hyzer, no ani, no nose up or down). Given that scenario, I'm pretty sure that once the disc is released, all that really matters is: Air speed and spin. Can someone tell me how each of these variables affect disc flight?

1) Two identical throws differ ONLY in air speed (same disc, same RPM, same throw, etc).
How would their flights differ?

I think discs tend to act "understable" at higher air speeds and "overstable" at lower airspeeds.

2) Now hold airspeed (and all other factors) constant and change rate of spin; two otherwise identical throws, one disc spinning faster than the other. How would their flights differ?

Not really sure how rate of spin affects flight, other than the disc needs to be spinning to achieve ANY stability. No Spin = pushing a disc = floppy short throw.

Inquiring minds want to know!

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Basic Physics say that object that spin as they move through the air travel faster as the rotation reduces the amount of resistence from the air.

So, all things told, a disc traveling at a faster speed should go further but there should be little to no difference in the flight pattern of the disc apart from the fade happening at a different point.

A disc with more RPMs will act more understable than its slower spinning counterpart not to mention it should travel faster and further. You will have to throw the disc harder to get it to travel at the same speed with less spin and still it will not travel as far.

The guys that I see throw it really far aften look like they are barely throwing a disc because it is all about getting the right snap on the disc to make it spin like crazy. Sure they are still probably getting more power and throwing the disc fast than I am as well but it is the spin they are really after.

I can give an in depth analysis of this later tonight... have too much work to get into it right now. I am a mechanical engineer specializing in the fluid dynamics of circulation lift theory. Saw another post on here that starts to explain some of what's going on with the disc, it has a few things wrong (i.e. what really causes lift), but most of it is pretty spot on. Here's the link to get you started:

I want to know this as well! Good question! And well phrased, too.

When I throw my putters or certain mids for longer shots(teeing off with them) I throw differently and usually get entirely different forms of flight when compared to my distance drivers. I suspected this (Disc Speed vs RPM) to be the cause. With my putters, I slow down a lot and usually don't accelerate until very late in the throw with a focus on a greater degree of wrist travel(for extra "snap"). These throws tend to be slow moving, but they seem to stay in flight longer and hold their line for a greater period of time. However, the decreased velocity of the disc causes the end result to not be all that much different from my driver.

On the other hand, when I throw my drivers, the throw is much longer and a hard pull is present throughout. In short, the velocity of the escaping disc is much higher, but the spin isn't as much. Thus, the disc reaches a good distance (325 is a pretty solid throw from me), but a strong fade is present. It is almost as if the disc is trying to fade much earlier(in terms of elapsed time) than the putters I mentioned, but it just covers much more ground in that alllowed time, be it smaller or not. Then again, I'm sure much of that is caused by the fact that a putter is inherently much, much straighter than a driver.

If all those complicated descriptions don't work, it is as though my putters(more spin, less speed) cruise, while my drivers are just thrown to their destinations.

With all that said, I would hypothesize from my above example that arm speed with decrease stability. Spin seems to do so as well, but I like to think of it as more of "maintaining" a discs stability. It's pretty damn hard to turn a disc over with just spin and no additional disc speed, but then again, no matter how fast one is moving, if it isn't spinning, it's not turning over. So...both?

Excellent question (but I don't know the answer). I have been procrastinating on posting a similar question, which is...

For the pros out there, do you make a conscious effort to change the ratio of spin to air speed from shot to shot? If so, in what situations?

For me, I either try to throw a disc harder or softer, with no real thought about isolating the two components, but I have a feeling pros think about it a lot more.

I think your question gets to the same point I'm curious about.

Excellent question (but I don't know the answer). I have been procrastinating on posting a similar question, which is...

For the pros out there, do you make a conscious effort to change the ratio of spin to air speed from shot to shot? If so, in what situations?

For me, I either try to throw a disc harder or softer, with no real thought about isolating the two components, but I have a feeling pros think about it a lot more.

I am no pro but I'll answer anyway. For the most part I do the same as you. There is one shot where I very occasionally try to impart more or less spin. On a spike or rainbow anhyser I'll try to put more spin on the disc if I want it to keep the angle I threw it all the way down. If I want it to start flattening out towards the end of the flight I'll spin it less.

Really looking forward to colodisc's response!

here's a quick and dirty analysis of the situation presented. I may not get through it all because I'm trying to catch a round with my buddy and he'll be here in a few minutes, so there may be a "to be continued"....

Case 1) same spin, different speed high speed stability analysis.
assumptions:
RHBH throw
disc 1 initial speed = 40
disc 2 initial speed = 50
initial spin speed measured at farthest point of disc from rotational axis= 10

case 1 disc1)

lift calculations for high speed stability:
left side of disc relative speed to air = 40+10 = 50
right side of disc relative speed to air = 40-10 = 30
ratio of left side speed to right side speed = 50/30 = 1.6 (the higher this number the more high speed turn due to lift factors)

rotational stability will also be a factor that is dependent on the polar moment of inertia about the rotational axis of the disc. If you've ever done the bicycle wheel suspended buy a string experiment, it is similar to this. The faster the disc is spinning, the more it wants to stay straight so the more force it takes to move it from it's initial placement. This will be greatly different for different discs because drivers have most of their mass located at a far distance from the rotational axis, and putter have a fairly uniform distribution of mass, thus drivers won't be as sensitive to these lift factors as a putter will. For this case it is not as important as it will be in case 2 where the rotational stability will different for the different shots.

case 1 disc 2 high speed stability analysis:
ift calculations for high speed stability:
left side of disc relative speed to air = 50+10 = 60
right side of disc relative speed to air = 50-10 = 40
ratio of left side speed to right side speed = 60/40 = 1.5 (the higher this number the more high speed turn due to lift factors)

This means at a fixed spin rate the slower flying disc (1) will have more high speed turn than the faster disc (2) mainly because the ratio of rotational speed to disc speed is higher in a slower disc.

case 1 low speed stability analysis: to be continued....

Colodisc - Hope you had a great round! Thanks for getting it started and including that link, and much thanks to Rameka too. A lot of work went into that post. Anyone interested here should check that out. (Especially Brother Dave's response - LOfreakin'L)!

Haven't digested all this info yet, but I'm working on it. The bit about the ratio of the velocities on the L & R sides of the disc affecting lift makes sense. I always knew the lift forces had to be different on the L & R sides because one side is traveling faster. Thanks for helping to clarify that.

still putting together my thoughts for the low speed fade portion of my analysis. There are a few different schools of thought on this phenomena, and I'm thinking that the magnus effect analysis will be easier to understand than precession due to vorticity for someone who doesn't study fluid mechanics, and also easier for me to explain without lecturing to much on viscosity, the no slip condition, turbulent flow, free stream velocity, starting vortex, circulation flow field, etc... gimme a few minutes to run some calculations and I'll decide if i'm taking the quick and easy explanation, or if I want to get into the real real flow analysis.

Where's Rameka when you actually need him?

Brother Dave, promise me you'll keep that gun tucked safely away! I don't want to be responsible for anyone's brains getting splattered all over their monitor.

I was reading through the thread Colodiscgolfer included and came up with a couple of conclusions:

1) There's a hell of a lot more to the dynamics of disc flight than I realized.
2) Some of you guys are freakin' hilarious.

I love science and math, but I love laughing even more. I'm simultaneously:
1) Digging the physics lessons
2) Laughing at OmegaSuperSloth's comments in Rameka's thread (keep 'em coming Omega).
3) Hoping Brother Dave doesn't put the gun to his head again
4) Glad I joined the DGCR community.
With any luck, I'll even get to play a round soon!

It's ALL good baby!

sorry for the delay guys, whiskey drinking commenced and I lost focus. Doesn't look like the magnus effect theory works here by my calculations... going to have to do it from a precession standpoint, and I just don't have the energy for it right now.

rome wasnt built in a day, take your time,

oh , practice practice practice

patiently awaiting your formulas, got the geek in me excited

Colo,

This is excellent stuff! Thanks so much for your clear explanation. Would you mind answering some questions I have on the "Physics of Flying discs" thread?

For others who are interested, here are some more articles on this subject. Look here and here.

Ok, I've run this over a few times and basically it comes down to the height at which the disc is at when the gravitational forces begin to take over the lift forces generated by the discs forward motion. A lot of assumptions need to be made to get a clear answer on the same spin different speed case. The original question poses the situation where the disc is thrown perfectly flat. This really isn't the best way to get the disc to go straight because of the misalignment of the center of lift pressure, and the center of mass. When folks say 'nose down', there's a reason. To actually get the disc to stay flat, without any change in elevation over the high speed portion of the flight, the pitch needs to be somewhere between -2, and -9 degrees from the horizontal depending on the aerodynamic properties of the disc.

I'll run two scenarios here, one where both discs are thrown at exactly the same height (actually impossible with the same spin and different speeds), and another where the faster disc is higher (more likely scenario) when the lift forces due to the speed are overcome by gravitational forces. Basically when the disc starts falling (gravity now stronger than lift), the disc itself will act as sort of a parachute because it now has a horizontal component to the velocity vector. The spin of the disc is still creating a lift pressure that is centered somewhere directly ahead of the disc's center of rotation, in the direction of travel. The amount of lift decreases in a clockwise direction around the disc when viewed from above (because of the RHBH spin direction). To visualize the lift vectors, picture the threads of a screw that has been cut in to a piece where there's one full revolution of the center post in the threads. The highest thread is at the front of the disc. (I know this isn't exact, but a good idea of the general picture we're looking at) This will cause the disc to turn more nose up because there is a lift force greater on the front of the disc than on the back, and to turn right because the lift pressure is greater on the right side of the disc than the left.

If we assume they are beginning to drop from the same height:
The higher speed disc will have lower spin velocity because it has traveled further, and thus more damping has occurred to the spin velocity. This will translate to a lower lift ratio both front to back and right to left than the slower thrown disc. This means that if it is falling from the same height, the higher initial speed disc will fade less at the end of it's flight than the slower speed disc.

If we assume the faster disc has gained more elevation than the slower disc:
It all comes down to how much the difference in height is. The faster disc will have further to fall, but the angle of turn is also decreased as in the first case. They may land in the exact same distance off to the left of the original line of throw, but the slower disc will turn harder, and doesn't have as far to fall. In this case there's too many other factors that come into play to really give a 100% dead sure answer unless we make a ton of other assumptions. Basically... the answer is 'it depends'. The big thing to take form this is that the faster disc will have a smaller degree of fade from the original direction, but will carry further in this fade.

Case 2 will be continued at a later time...

Excellent question (but I don't know the answer). I have been procrastinating on posting a similar question, which is...

For the pros out there, do you make a conscious effort to change the ratio of spin to air speed from shot to shot? If so, in what situations?

For me, I either try to throw a disc harder or softer, with no real thought about isolating the two components, but I have a feeling pros think about it a lot more.

I think your question gets to the same point I'm curious about.
I'm not a pro, but my understanding is that for drives you don't try to control spin. You control how hard you throw by limiting your reach back or taking fewer (or no) steps, but there isn't anything you can do, or any reason, to consiously try to control spin at those speeds and with that technique.

At approach speeds I can see a benefit to controlling spin. It's also a lot easier to do. Sometimes you need more or less carry than you'd normally get so you can add or take spin away for shorter shots. It's not something you'll need to do a lot, but I can see how you could control it.

I'll reiterate that timing and mechanics are what you should be focusing on to improve your drive.

To actually get the disc to stay flat, without any change in elevation over the high speed portion of the flight, the pitch needs to be somewhere between -2, and -9 degrees from the horizontal depending on the aerodynamic properties of the disc.

The big thing to take form this is that the faster disc will have a smaller degree of fade from the original direction, but will carry further in this fade.

Awesome stuff! I've been wondering about the nose down angle for a long time. Can you say that the -9 degrees would be more for a blunter nosed disc like a fairway driver or is that backwards or are there more variables involved like domeyness of the flight plate?

The big take-away fits with how I shoot shorter shots that need to turn left. I'll generally use a higher speed disc, throw it slower and let it hook left hard.

-9 would be for a deep rimmed putter like a birdie or rec disc, -2 would be for something more in the range of a high speed driver like a boss or force

Ok, I've run this over a few times and basically it comes down to the height at which the disc is at when the gravitational forces begin to take over the lift forces generated by the discs forward motion. A lot of assumptions need to be made to get a clear answer on the same spin different speed case. The original question poses the situation where the disc is thrown perfectly flat. This really isn't the best way to get the disc to go straight because of the misalignment of the center of lift pressure, and the center of mass. When folks say 'nose down', there's a reason. To actually get the disc to stay flat, without any change in elevation over the high speed portion of the flight, the pitch needs to be somewhere between -2, and -9 degrees from the horizontal depending on the aerodynamic properties of the disc.

I'll run two scenarios here, one where both discs are thrown at exactly the same height (actually impossible with the same spin and different speeds), and another where the faster disc is higher (more likely scenario) when the lift forces due to the speed are overcome by gravitational forces. Basically when the disc starts falling (gravity now stronger than lift), the disc itself will act as sort of a parachute because it now has a horizontal component to the velocity vector. The spin of the disc is still creating a lift pressure that is centered somewhere directly ahead of the disc's center of rotation, in the direction of travel. The amount of lift decreases in a clockwise direction around the disc when viewed from above (because of the RHBH spin direction). To visualize the lift vectors, picture the threads of a screw that has been cut in to a piece where there's one full revolution of the center post in the threads. The highest thread is at the front of the disc. (I know this isn't exact, but a good idea of the general picture we're looking at) This will cause the disc to turn more nose up because there is a lift force greater on the front of the disc than on the back, and to turn right because the lift pressure is greater on the right side of the disc than the left.

If we assume they are beginning to drop from the same height:
The higher speed disc will have lower spin velocity because it has traveled further, and thus more damping has occurred to the spin velocity. This will translate to a lower lift ratio both front to back and right to left than the slower thrown disc. This means that if it is falling from the same height, the higher initial speed disc will fade less at the end of it's flight than the slower speed disc.

If we assume the faster disc has gained more elevation than the slower disc:
It all comes down to how much the difference in height is. The faster disc will have further to fall, but the angle of turn is also decreased as in the first case. They may land in the exact same distance off to the left of the original line of throw, but the slower disc will turn harder, and doesn't have as far to fall. In this case there's too many other factors that come into play to really give a 100% dead sure answer unless we make a ton of other assumptions. Basically... the answer is 'it depends'. The big thing to take form this is that the faster disc will have a smaller degree of fade from the original direction, but will carry further in this fade.

Case 2 will be continued at a later time...

Oh boy, the room is spinning again, where's my dang revolver at? Stupid Christopher Walken, never returning my stuff.

Appreciate the help!

First things first - avoiding tragedy:
Brother Dave, it's gonna be OK. Just hand me the gun, niiiice 'n' eeaasy . Thaat's it. Now go to the fridge, get yourself a cold one, pop the top and ease back down onto the sofa. Take a good long sip. Isn't that better?
But when I have a history question, you'll be the guy.

Colodiscgolfer, THANKS FOR THE ANALYSIS! After reading the thread you referred me to, I realized there's no simple answer but...

The original question poses the situation where the disc is thrown perfectly flat. This really isn't the best way to get the disc to go straight because of the misalignment of the center of lift pressure, and the center of mass. When folks say 'nose down', there's a reason. To actually get the disc to stay flat, without any change in elevation over the high speed portion of the flight, the pitch needs to be somewhere between -2, and -9 degrees from the horizontal depending on the aerodynamic properties of the disc.

I'll run two scenarios here, one where both discs are thrown at exactly the same height (actually impossible with the same spin and different speeds), and another where the faster disc is higher (more likely scenario) when the lift forces due to the speed are overcome by gravitational forces.

Totally makes sense, as the speed should affect lift. I suppopse the air pressure is greater towards the front of the disc than it is at the back, which means lift force would be further forward than the center of mass - is this correct?

If we assume they are beginning to drop from the same height:
The higher speed disc will have lower spin velocity because it has traveled further, and thus more damping has occurred to the spin velocity. This will translate to a lower lift ratio both front to back and right to left than the slower thrown disc. This means that if it is falling from the same height, the higher initial speed disc will fade less at the end of it's flight than the slower speed disc.

If we assume the faster disc has gained more elevation than the slower disc:
It all comes down to how much the difference in height is. The faster disc will have further to fall, but the angle of turn is also decreased as in the first case. They may land in the exact same distance off to the left of the original line of throw, but the slower disc will turn harder, and doesn't have as far to fall. In this case there's too many other factors that come into play to really give a 100% dead sure answer unless we make a ton of other assumptions. Basically... the answer is 'it depends'. The big thing to take form this is that the faster disc will have a smaller degree of fade from the original direction, but will carry further in this fade.

Trying to comprehend this: As gravity overcomes lift, the disc begins to fall. If we break each disc's velocity down into component vectors, the faster disc's forward vector is obviously greater, but their leftward component vectors would be fairly similar (I know you made some assumptions), so they appear to fade, as they fall due to gravity with similar leftward components of their velocity vectors. The less "fairway" that is covered during the fade, the harder the fade appears to be. Thus, the disc that's thrown slower appears fade harder. This may be more a human observation, than a mathematical description.

In reality, they travel (fade) similar distances to the left, but
a) the fade begins at different distances down the fairway, depending on initial velocity.
b) the faster disc travels further (down the fairway) after fade begins, so it's path is less curved, even though it likely ends up about as far to the left.

Does that hold water?

Also, is the leftward motion during the end of flight due (at least in part) to precession? I know that wasn't part of the original question, so if the answer's too involved to cover here, I can live. Might be able to find it somewhere, just thought I'd ask cause I think its starting to gell.

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