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Project Pothole


yotaman88210

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We are currently working on something, and need the help of all our members. This is being posted in State-Wide for that very reason.

Im not sure about all of you, but I have seen WAAAY too many potholes while driving around Columbus. All I can think when I see/hit these is "thank god im not on the bike!" I have seen a few holes lately that would easily throw me off my bike, or damage the bike. One of our members hit a hole last season and damaged his wheels. What we are trying to get together is a list of approximate location and if possible(and safe) a photo of the potholes in Ohio. I know most of our members are in the greater Columbus area, but we do have some people in other areas. We really need everyones help in doing this. The ultimate Goal is for these said holes to be repaired by the time its warm enough to ride everyday.

So.... how do you help. Keep a notebook and pen in your vehicle and write down where you are when you see a damaged road. If the damage happens to be somewhere where you are at a standstill, i.e. near traffic lights or stop signs, or on the highway during rush hour, try to get a quick photo of it. If we can get a few good photos of the larger potholes, I feel we will have a much better fighting chance of getting the holes repaired FAST!

I will be contacting the city to get these damaged roads repaired once we have a nice list. THANKS!

LIST

Tamarack and the service road that's parallel to 161

Tamarack just before Brookfield

I-71 north between the merge from 670 east -- the slow lane but not the merge lane

south old state and lazelle

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The engineer in me is dying to know... If the suspensions can handle coming down on wheelies and doing endos and whatnot, how big do these potholes have to be before actual DAMAGE results? Just curious...

You think it's more likely you'll get tossed from the bike due to the pothole before it actually breaks something?

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The engineer in me is dying to know... If the suspensions can handle coming down on wheelies and doing endos and whatnot, how big do these potholes have to be before actual DAMAGE results? Just curious...

You think it's more likely you'll get tossed from the bike due to the pothole before it actually breaks something?

The engineer in you should realize that hitting a pothole at 70mph is going to involve more force then your front tire hitting the ground from a wheelie from 4-5 ft in the air.

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Not to mention the angle of the forces involved, when you come down from a wheelie, the bike is moving straight and coming down generally perfectly on the suspention. Hitting a pot hole, the wheel is slammed into it, like a curb, which is 90* out of phase from the travel of the suspention. If it's large enough, it will hit high up on the rim, more towards the centerline of it, and put more of a rearward, than upward force on the suspention. Wheels are obviously round, and from 0* (being the bottom, on the ground) to 45* up on the front, any objects will be transmitted upwards, after that 45* mark the force gets exponentially applied rearward, not upward.

What makes potholes so bad is that not only is the bike now leaning forward with downward/forward velocity, the lip of the pothole is further up on the wheel than it would be if you hit that same 3" curb going flat, because of the downward angle of the bike.

Thats about as technical as I cared to make it. I could get more involved if you like :D

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Not to mention the angle of the forces involved, when you come down from a wheelie, the bike is moving straight and coming down generally perfectly on the suspention. Hitting a pot hole, the wheel is slammed into it, like a curb, which is 90* out of phase from the travel of the suspention. If it's large enough, it will hit high up on the rim, more towards the centerline of it, and put more of a rearward, than upward force on the suspention. Wheels are obviously round, and from 0* (being the bottom, on the ground) to 45* up on the front, any objects will be transmitted upwards, after that 45* mark the force gets exponentially applied rearward, not upward.

What makes potholes so bad is that not only is the bike now leaning forward with downward/forward velocity, the lip of the pothole is further up on the wheel than it would be if you hit that same 3" curb going flat, because of the downward angle of the bike.

Thats about as technical as I cared to make it. I could get more involved if you like :D

+1 my point exactly

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there's a nice pothole at the corner of south old state and lazelle in lewis center/columbus.

do I have to get a picture of it for this scheme to work?

Yea, I hit that one everyday on the way to work. Its a biggen. A pic would be nice but I will prolly get one of it.

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Thats about as technical as I cared to make it. I could get more involved if you like :D

I agree with all that, but I was wondering if anyone had any hard numbers or spec sheets for forks -- I can't find any (admittedly I haven't looked that hard), but I suppose that the hard numbers lie as classified information inside the manufacturers of the bike and suppliers of the forks. Given the second-order nature of hydraulically damped suspension, I'd even settle for a Force vs. Displacement curve.

Free-body diagram time?

Using the 'busa as an example, which has a 24º rake. If the bike is 550lbs with fluids and you assume the claimed 49R/51F weight distribution = 281lbs on the front wheel at the "zero" point, which at 24º rake is 307lbs [281/cos(24º)] total force (281lbs normal force, and 26lbs horizontal force holding the tire in front of the bike).

With a rider, mass becomes 383lbs at the front wheel = 419lbs total force.

Rim + Tire size = 20.3" diameter [17" rims + 120mm*0.7 aspect ratio then convert to inches]

If the pothole is 3" deep, the angle of attack becomes 31º above the ground plane (according to AutoCAD); although the bike won't be level which will reduce the rake ~ 3º over the wheelbase = 21º

If the pothole is 6" deep, the angle becomes 44º; yet again, the 6" will affect the 'level' of the bike and decrease the rake effectively ~ 6º = 18º

Of course you have to take into account the weight distribution has shifted (Front tire in pot hole will load the bike more on the front, when the rear tire is on "level" ground)

At 70mph, the bike has a momentum of 750lbs*70mph = 52500 lbs*mph (p=mv). Now let's assume the front tire is in the 6" pothole, and the weight ratio is now 30R/70F; therefore the front tire has momentum of 36750 lbs*mph and it will be looking to maintain momentum throughout the pothole event.

The bike must cover 1209600 lbs*in/s2 (front wheel gravity force over 6" vertical) in a distance of 8.67 in @ 70mph (1231.92 in/s) or in 0.007seconds.

= 8513 lbs*in/s (upward force); and since I don't know the spring rate / dampening, I can't solve for the lbs force

= 8816 lbs*in/s (horizontal force); same as above

And now, it gets way more complicated, and my AutoCAD is getting squiggles everywhere, so I stopped. :cool:

(And I'm sure I messed up somewhere...check my work, please!)

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The engineer in me is dying to know... If the suspensions can handle coming down on wheelies and doing endos and whatnot, how big do these potholes have to be before actual DAMAGE results? Just curious...

You think it's more likely you'll get tossed from the bike due to the pothole before it actually breaks something?

Actually, the suspension and frames on today's sportbikes DONT handle coming down from wheelies that well. I've seen plenty of blown fork seals, broken wheels, and bent frames from a "few wheelies".

The engineer in you should realize that hitting a pothole at 70mph is going to involve more force then your front tire hitting the ground from a wheelie from 4-5 ft in the air.

Maybe he meant the kind that drives the train?? :D

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if you wanna get em fixed, start making phone calls. IIRC, they wont even think about coming out to look at it unless they get like four phone calls.

everyone needs to call. and call like three times a day. this is the government you are dealing with, so you gotta keep em on track, and most importantly, dont expect anything to happen fast.

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I agree with all that, but I was wondering if anyone had any hard numbers or spec sheets for forks -- I can't find any (admittedly I haven't looked that hard), but I suppose that the hard numbers lie as classified information inside the manufacturers of the bike and suppliers of the forks. Given the second-order nature of hydraulically damped suspension, I'd even settle for a Force vs. Displacement curve.

Free-body diagram time?

Using the 'busa as an example, which has a 24º rake. If the bike is 550lbs with fluids and you assume the claimed 49R/51F weight distribution = 281lbs on the front wheel at the "zero" point, which at 24º rake is 307lbs [281/cos(24º)] total force (281lbs normal force, and 26lbs horizontal force holding the tire in front of the bike).

With a rider, mass becomes 383lbs at the front wheel = 419lbs total force.

Rim + Tire size = 20.3" diameter [17" rims + 120mm*0.7 aspect ratio then convert to inches]

If the pothole is 3" deep, the angle of attack becomes 31º above the ground plane (according to AutoCAD); although the bike won't be level which will reduce the rake ~ 3º over the wheelbase = 21º

If the pothole is 6" deep, the angle becomes 44º; yet again, the 6" will affect the 'level' of the bike and decrease the rake effectively ~ 6º = 18º

Of course you have to take into account the weight distribution has shifted (Front tire in pot hole will load the bike more on the front, when the rear tire is on "level" ground)

At 70mph, the bike has a momentum of 750lbs*70mph = 52500 lbs*mph (p=mv). Now let's assume the front tire is in the 6" pothole, and the weight ratio is now 30R/70F; therefore the front tire has momentum of 36750 lbs*mph and it will be looking to maintain momentum throughout the pothole event.

The bike must cover 1209600 lbs*in/s2 (front wheel gravity force over 6" vertical) in a distance of 8.67 in @ 70mph (1231.92 in/s) or in 0.007seconds.

= 8513 lbs*in/s (upward force); and since I don't know the spring rate / dampening, I can't solve for the lbs force

= 8816 lbs*in/s (horizontal force); same as above

And now, it gets way more complicated, and my AutoCAD is getting squiggles everywhere, so I stopped. :cool:

(And I'm sure I messed up somewhere...check my work, please!)

:???::dunno::wtf:

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I agree with all that, but I was wondering if anyone had any hard numbers or spec sheets for forks -- I can't find any (admittedly I haven't looked that hard), but I suppose that the hard numbers lie as classified information inside the manufacturers of the bike and suppliers of the forks. Given the second-order nature of hydraulically damped suspension, I'd even settle for a Force vs. Displacement curve.

Free-body diagram time?

Using the 'busa as an example, which has a 24º rake. If the bike is 550lbs with fluids and you assume the claimed 49R/51F weight distribution = 281lbs on the front wheel at the "zero" point, which at 24º rake is 307lbs [281/cos(24º)] total force (281lbs normal force, and 26lbs horizontal force holding the tire in front of the bike).

With a rider, mass becomes 383lbs at the front wheel = 419lbs total force.

Rim + Tire size = 20.3" diameter [17" rims + 120mm*0.7 aspect ratio then convert to inches]

If the pothole is 3" deep, the angle of attack becomes 31º above the ground plane (according to AutoCAD); although the bike won't be level which will reduce the rake ~ 3º over the wheelbase = 21º

If the pothole is 6" deep, the angle becomes 44º; yet again, the 6" will affect the 'level' of the bike and decrease the rake effectively ~ 6º = 18º

Of course you have to take into account the weight distribution has shifted (Front tire in pot hole will load the bike more on the front, when the rear tire is on "level" ground)

At 70mph, the bike has a momentum of 750lbs*70mph = 52500 lbs*mph (p=mv). Now let's assume the front tire is in the 6" pothole, and the weight ratio is now 30R/70F; therefore the front tire has momentum of 36750 lbs*mph and it will be looking to maintain momentum throughout the pothole event.

The bike must cover 1209600 lbs*in/s2 (front wheel gravity force over 6" vertical) in a distance of 8.67 in @ 70mph (1231.92 in/s) or in 0.007seconds.

= 8513 lbs*in/s (upward force); and since I don't know the spring rate / dampening, I can't solve for the lbs force

= 8816 lbs*in/s (horizontal force); same as above

And now, it gets way more complicated, and my AutoCAD is getting squiggles everywhere, so I stopped. :cool:

(And I'm sure I messed up somewhere...check my work, please!)

:lol: So did you get anything else accomplished today? :cheers:

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