Got Math?

jmc

jmc

Joined
May 23, 2008
Messages
11
Location
Monument, CO
Bike
'01 ST1100A
STOC #
5198
This question occurred to me as I was trying to pick it up: how much weight am I lifting?

This question is waaayy beyond my Math skills, but I hope someone out there will be able to work the problem out.
 
Probably a lot more than you would like, but a lot less than you think...... as long as you use the correct method it isn't too difficult... DAMHIK.
 
Hey Jim ... depends on position of the CG of the bike (vertically) and where you grab it when you try to lift it back up. Very roughly, if the bike were horizontal, at the start of the lift the tires would carry half the weight and you'd be lifting the other half... but the bike isn't horizontal when you start the lift.

I'd guess about 40% of the bike's weight at most, more likely ~30% or less.
:shrug1:
 
Military press, clean jerk or bench press?

Roughly:
Bike weighs ~635lbs dry and unloaded.
1 gal water
1 gal oil
7 gal of gas.
9gal @ ~8lb/gal = 72lbs

So you're working with around 707lbs if you don't have anything in your bags or pockets and have no other changes.
 
Not enough info. To pick it up on a truck, off the ground or to get it on centerstand?
 
Karen, I'm pretty sure the bike weighs the same either way although why it fell over in the 1st place is another story LOL.
 
Actually, I'd say just the opposite. No one around? #$)(#@*&$@)#*&% bike weighs 9000 lbs..

A team of college age cheer leaders and their sexy coach? Bike would be upright before anyone saw it drop. 0 lbs. :D
 
Mark said:
Military press, clean jerk or bench press?

Roughly:
Bike weighs ~635lbs dry and unloaded.
1 gal water
1 gal oil
7 gal of gas.
9gal @ ~8lb/gal = 72lbs

So you're working with around 707lbs if you don't have anything in your bags or pockets and have no other changes.
Click to expand...

Mark: Probably eluding to how much force is required. You'd have to figure out location of CG vs wheels as fulcrum and where he is grabbing the bike for leverage.
 
The answer is pretty easy to get if you know how high the center of gravity of the bike is from the ground when it is sitting level. With that information and the angle the bike is setting you can determine the lifting force required.

Based on nothing more that a gut feel, I'd guess you are picking up a little less than a third of the bike's weight- or using the numbers above, around 200 pounds. My memory of lifting mine would seem to confirm that's about what it felt like.

The first time I picked mine up (on a lonely country road with no one around in the middle of the night), I had to unload the camping gear before I could budge it. Because of the lift points, anything on the bike contributes DIRECTLY to the lifting force required. (OK- if there are any other engineers reading this, I know that "Directly" isn't quite right- but it is pretty close).

If someone knows where the CG is located, I'd be happy to take a measuring tape and figure the exact force required. Either that or I can get a BIG FISH scale, a couple of pulleys, some rope and a sturdy tree......
 
Thanks for the replies, guys. Here's how I think of it. If we assume (unrealistically) for a moment that that the CG is at the very top of the bike, the weight to lift is about 700 lbs. If the CG is at the bottom of the tires (also way unrealistic), the weight to lift would be around zero lbs. If BakerBoy's low estimate (30%) is close, then I'm trying to lift 210 lbs (30% of 700), which ain't easy.

I know the correct way to do it is with your butt and your legs, but if the road is sandy your feet can't get enough purchase to do it.

How about a skier's airbag that inflates and lifts the bike part-way. Or lift jacks like on an Indy car. Or a sky-hook . . . :)

Thanks again for the help.
 
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