| Author | Message | ||
     Davegess |
And Blake and Greg, I am still waiitn to see the same sort of info on turning a side car ;o> what's the hold up, it's not that they are more complicated than a regular bike;o) | ||
| Joey |
Davegess! Before I got my windshield, my forearms were sore every day from riding 50+ miles at 70 MPH twice a day. My stomach muscles were getting a workout, too. | ||
| Sparky |
Sidecars, especially unladen rigs, have a real problem making right turns safely without flying the chair and going out of control. Normally sidecarists depend on adding ballast or a passenger to keep the chair on the ground. People that don't have gotten killed. But what if you want a lightweight performance bike with a lightweight sidecar? You're out of luck unless you add about a 100 lbs ballast in the car or a passenger. Then performance goes out the window. Maybe the answer is to add a lightweight flywheel in the sidecar to generate enough gyroscopic force to keep the car from lifting. Waddya think, can technology save the day? BTW I saw Steve Wozniak (Apple co-founder) on TechTV last night cruising around the show on a Segway. How's that for a practical application of gyro-tech.  Sparky, ex Equalean pilot \-\ /-/ | ||
| Blake |
Just learned this the other day... The Segway does not use an actual mechanical gyro, it uses sensitive instrumentation to control the servo motors driving each wheel. They call it a virtual gyro or some such thing. | ||
| Sparky |
Hmmm... I did notice that the platform he was standing on was below the axle height by an inch or 2. Sorta self-centering then, like standing on the swing seat of a swing set on wheels, sorta. | ||
| Dynarider |
Sparky, I would think sidecars could take a right turn well, its the left handers that are a bitch. Had Rick out of Hals use his sidecar outfit on the battletrax course last yr & the righthanders he snapped right thru. | ||
| S320002 |
Sorry Dave, way more variables than I want to deal with there. 'Sides I don't know which way you want to turn. ;-) Greg | ||
| Blake |
Dave, I give up. I hate sidecars anyway.  | ||
| Sparky |
Dyna, It's not as intuitive as it seems: For street rigs the SC is used as an outrigger to prevent overturn in left handers but in right turns the weight of the SC plus the driver (& passenger) leaning into the turn keeps the rig from tipping (hopefully). Racing rigs are different because they can slide around. And you're correct -- a rig can potentially make right handers faster because the bike essentially drives around the SC whereas in lefties the bike pulls the SC a greater distance. Ask Rick how much ballast he used to keep the SC down. Sparky | ||
| S320002 |
Blake, I've done the slalom thing hundreds of times. Maybe it's just me, but the steering has never felt very difficult. Greg | ||
| José_Quiñones |
Try slaloming using different gears while maintaining a constant 35-40 mph, see if you notice the third "gyro" as the engine rpm varies. | ||
| Blake |
Greg, If you have never noticed how difficult high speed aggressive steering can be, I can guarantee it is you. If you don't try to turn aggressively, you won't notice the effect. Basically in the vacant highway slalom exercise (perform at your own peril, it is not recommended by me or anyone ![:]](http://www.badweatherbikers.com/buell/clipart/proud.gif) ) you end up trying to go from full lean left to full lean right as hard and quicly as possible, you want to execute the transition more quickly, but the bike simply will not allow it and you are about bending the handlebars from pushing/pulling on them so hard. JQ, It may be my imagination but I seem to recall that entering turns 3 and 6 at OHR in 3rd versus 2nd does result in a slightly easier turn in. My entry speed is at around 45 mph. I found it better to just leave it in 3rd, not for the easier turn in, just to avoid having to upshift while leaned over and going uphill. | ||
| Nevco1 |
And Blake Says..."It may be my imagination but I seem to recall that entering turns 3 and 6 at OHR in 3rd versus 2nd does result in a slightly easier turn in. My entry speed is at around 45 mph. I found it better to just leave it in 3rd, not for the easier turn in, just to avoid having to upshift while leaned over and going uphill." NOW THAT'S THE BEAUTY OF A BUELL!!! | ||
| Blake |
Greg, If I had one, I'd send you one of those briefcase gyros. For the dynamically inclined... Let's do some comparing of the different forces at work and their relative effects on the bike/rider. To start let's compare the total kinetic energy stored in angular momentum of the wheels versus linear momentum of the bike/rider. For the linear case... So at 50 m/s (112 mph) with a 300 Kg (660 LB) bike/rider combination we get... For the angular (rotational) case... Wheel rotational speeds at 50 m/s are approximately 60 rad/s (576 rpm)and 63 rad/s (602 rpm) for the front and rear, respectively. Front and rear wheel assemblies moments of inertia are estimated to be (Io) to be 0.4 and 0.6 Kg-m2, respectively. So the total kinetic energy stored in the wheels at 112 mph (50 m/s) is... Summing linear and angular we find the total kinetic energy in round numbers is 278,000 FT*LB. How does that compare to the roll rate related kinetic energy of a typical sport bike with rider when rolling/leaning at 180°/s? Estimating the roll axis moment of inertia of the bike to be approximately 6.5 Kg*m2 and finding that 180°/s is equal to 3.142 rad/s (there are p=3.142 radians in 180 degrees) we find the kinetic energy of such a roll/lean rate to be... To achieve that roll rate within a half second would require how much power and how much torque? Since power (P) is equal to energy (E) divided by time interval (t)... (recall that 1.0 HP = 550 FT*LB/s). What torque is required to generate that amount of power in 1/2 second? Since we know that power is also equal to torque (T) times angular speed (w) or... we can find the torque as... As a check we also know that torque equals mass moment of inertia (I) times angular acceleration (w') or... Angular acceleration is equal to the delta (change in) angular velocity divided by the time interval (t). The initial w is 0 rad/s with a final w of p rad/s, so we have... the same as the result obtained using the energy method. It's good when things agree. Next let's find the moment (M) trying to tip a bike/rider with a center of mass about 30 inches above the pavement when vertical. It depends on how far the bike is leaned; no lean, no tipping moment. If the bike is leaned at 45°, the moment will be equal to the weight of the bike/rider, say 660 LB, times the horizontal distance from center of mass to center of contact patch. We roughly estimate that horizontal distance using good old trigonometry as SIN(45°)*30" = 21.2" = 1.767'. So the tipping moment due to gravity and lean will be 660 LB * 1.767' = 1,167 FT*LB Now let's find what radius curve taken at 50 m/s (112 mph) and a lean angle of 45° would exactly offset that tipping moment. The centripetal force acts at the contact patches pushing inwards towards the inside of the curve. That force causes a restoring moment (untipping moment) acting through the moment arm defined as the vertical distance from contact patch to center of mass. Since the bike under consideration is at 45°, the vertical distance from pavement to contact patch is virtually identical to the horizontal distance. Anyway it is close enough for this investigation. When in equilibrium (constant lean angle) the tipping moment must exactly equal the restoring moment. Since the restoring moment arm is the same as the tipping moment arm, and the bike/rider are in equilibrium maintaining a constant lean angle, then the centripetal force must exactly equal the gravitation force (weight). So the centripetal acceleration is exactly 1.0 g and the centripetal force is 660 LBs the exact same as the bike's weight. Now you know why I chose a 45° lean angle. So anyway we want to find the radius of a curve that will produce a 1.0 g centripetal acceleration at 50 m/s (118 mph). It just so happens that... where "r" is the radius of the curve. Solving for the radius give us... So we can find that the radius of the curve would need to be... Now let's find the angular velocity of the bike/rider as they negotiate such a curve. If the curve were a complete circle, its circumference (c) would be equal to its radius times 2p (c=2pr), so we find... Traveling at 50 m/s we would cover 1,627m in time (t) of 32.5s (1,627m/50m/s=32.5s). A complete circle has 360 degrees which is equal to 2p radians, so our angular velocity w going around the curve would be equal to... Do the relative magnitudes of any of these estimates tell us anything? Anyone? | ||
| José_Quiñones |
you have WAY too much time on your hands.... | ||
| Hans |
Yes Blake it does: It makes clear why racers go faster around with heavy braking on the straight, then making a very sharp turn at very low speed and then accelerate again, turning very slightly the last part, instead of wandering around with the same, moderate, speed. Because the lean angle is already 45 degrees for such a very wide turn at that far below racing speed. Hans | ||
| S320002 |
Blake, I'm familiar with briefcase gyros. It would be more accurate to compare the briefcase handle with trying to steer your M2 by grasping the handlebar, with one hand, just above the steering pivot. Nearly impossible at any speed. Greg | ||
| Sparky |
Would such a briefcase gyro, if mounted horizontally to a bike, have enough force to prevent it from leaning in turns... i.e. corner like a sidecar? | ||
| S320002 |
No. | ||
| Andrewb |
Any way to get that on a T-shirt? With a BWB logo as well?? | ||
| Joey |
Blake! I have a question that has been nagging me for a while now. I'm hoping your physics background can help. I ride the Blast! to and from work in freezing weather quite a bit. Rather than come up with some protection for my fingers and toes, I'm thinking that, at some speed, the friction of the atmosphere can overcome wind chill. So, how fast would I have to go to warm my fingers up to 75 degrees when it is 25 degrees outside? And, how much horsepower (pounds of thrust?) would it take to achieve this speed? I know--it's unlikely that I could modify my Blast! to go the speeds I'm requesting, but I can dream, can't I? | ||
| Blake |
heheheh... Somewhere above Mach 2 might start to warm things up. | ||
| Aesquire |
I don't have the real numbers handy (sorry Blake) about 400mph you get real friction heating (and skin may not like that speed) Mach 2.2 is about the limit for aluminum airframes (Concorde) before you heat the skin enough to lose hardness (annealling) not melting the aluminum, but getting it soft, losing the heat treating. Ti is good out to 2500-2800 mph, the Blackbird re-heattreats each flight. 600 degrees +...  If anyone cares I'll haul out the NASA tech briefs. On your Blast? Maybeee Nitrous? Jato bottle?  | ||
| Aesquire |
After checking with NASA, maybe 350mph, depends on humidity, I'd suggest Gloves. Blast + Jato = Pink Mist. | ||
| Joey |
I wonder if the "Express" toll lane on highway 1 will read the tag at those speeds ... | ||
| Jima4media |
Centripetal Force - Misunderstood? Hi, I'm back again. My Directv DSL account went away when they went out of bidness, and now I'm back to AOL snail e-mail at 28K. The new SBC DSL gets installed this weekend. Anyway, centripetal force - it's what holds all the parts in the universe together, all the parts in your Buell together, and holds your wheels on the ground. If there is a god in the universe, it's centripetal force, sometimes better known as gravity. May the force be with you... Jim X-X.2.5 - Another story | ||
| Blake |
Jim, Your postscript reminds me. We need to add your story to the BadWeB Hall of Fame. Wanna tell it again? Centripetal force = gravity? No wonder your X1's have had such trouble. Greg, Check out the following marketing speal from Ducati's 999R "Features" page
Thought you might be interested. | ||
| S320002 |
Blake, That concept was touched on awhile back. It was in reference to the ZTL brake and the XB specific tires. In systems of equal mass, concentrating the mass toward the circumference increases the gyro effect. The XB front wheel/brake assembly is light but has more of its relative mass at the outer circumference. The lighter tire helps to reduce the effect. Maybe you missed it. Greg | ||
| Jima4media |
Blake, http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Specifically, http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf and, http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html#grav Unless, of course, magic gyros and voodoo make the heavens and earth move. Jim | ||
| Dynarider |
Where the hell did Jims post go? |
