How to weigh a bike......The weigh station is open, aw crap, closed for now.

[Moto]

If memory serves me right.  Chuckie post the pix of the tee.

Just describe it for us.

[Tom]

Nah....I'll let Chuckie post the pix.

[Moto]

Nah....I'll let Chuckie post the pix.

Pretty coy. You a teenage girl?

Say what's on your mind.

[Tom]

Nah....I'm  laptop set on auto-response.  I'm sitting on my owner's coffee table.  You a girl??

[Moto]

Nah....I'm  laptop set on auto-response.  I'm sitting on my owner's coffee table.  You a girl??

You're a troll.

[Tom]

Now....now....don't get your panties in a wedgie.  Take some time to take them out of your butt crack and take a chill pill.  Maybe you'd like to stop looking at yourself in the mirror.

[jdgretz]

Someone post a pix of the  Mr. Nevada tee.   

I'm sure you are asking for the one that says "For every problem there is one unexpected solution" rather than the one that says "I have always relied on the kindness of strangers"

jdg

[Moto]

I'm sure you are asking for the one that says "For every problem there is one unexpected solution" rather than the one that says "I have always relied on the kindness of strangers"

jdg

Ah. Well, in that case, never mind. I'll cancel the Vendetta Special tickets I bought on Alitalia last night.

Mr. Nevada is known to me only as the butt of countless character-demeaning references on this site. I supposed I was being called a fat, doughnut-gorging nincompoop. Which is ok if the one doing the calling says what he means. Context is important. Though Tom will never get an Emily Post award on my recommendation, I guess we can stop the Pee Wee Herman exchanges -- "I know you are, but what am I?" -- if that is what he was intending.

I thought I would go back to the original question while waiting for a clarification of the intended message, so this morning I plotted the implications of my two ideas of how to model the problem, and realized a basic error, which I will now explain, for any who may be revisiting this thread for something other than low comedy.

My first idea was to model the motorcycle -- remember the motorcycle? -- as a point mass on a weightless stick pivoting on the rear axle. With this model it is true that the weight measured at the mass is proportional to the cosine of the angle the stick makes with respect to  the horizontal. (This corresponds, of course, to the line drawn through the axles of the motorcycle.) The formula for predicting the weight in this model is weight1 = weight0 * cos(alpha), where alpha is the angle from the horizontal, and weight0 is the measured weight at zero elevation of the front wheel.

After a first set of hasty measurements that disconfirmed model 1 I went back to the lab (my workshop) and did much more careful measurements on the second test subject (shown above). I measured the wheelbase and the position of the front axle when the bike was balanced fore-and-aft by using a very lightweight plumb bob made of thread and a small eye screw, and taking my measurements over to the side of the bench with a try square. To weigh the front I used an electronic postage scale that I believe is accurate, and carefully measured the height of the scale's platform, to the nearest millimeter. I found it hard to get a reliable weight measurement, mostly because the bike's handlebar could rotate, so I took three measurements at each elevation setting. These are the data that I showed in the message above.

This graph shows the more accurate measurements on the second subject (i.e., experiment 2) as the set of three dots per elevation setting (nearly indistinguishable in some cases). The red dashed line is the model 1 prediction, which clearly doesn't fit the data.



I realized as I did the first experiment that my second proposition -- that the bike would balance perfectly fore and aft when rotated 90 degrees from horizontal -- was false. In fact, it balanced when the bike was rotated about 70.9 degrees from horizontal, because the center of mass was higher than the line drawn between the two axles. So, in effect, the bike was rotated 19.1 degrees further toward its balancing point than it appeared at any given elevation. Adjusting the basic cosine formula for this gave model 2: weight2 = weight0 * cos(alpha + 19.1). Model 2's prediction is shown as the blue dashed line on the plot. It doesn't fit either.

Only after doing all this -- and, I confess, entertaining myself along the way -- did it occur to me that I knew the reason why my procedures wouldn't work. The mechanical concept of a "moment," understood as an integral, is what's needed. The mass of the motorcycle is not really a point, but is distributed along its length. Because the rotational torque generated by the weight is a function of distance from the axle, not all weight can be treated equally in summing up the force. Instead, using f(x) to designate the point mass at any point x, as measured from the axle, the needed calculation for the weight seen at the scale is, I believe:

      \int cos(alpha) x f(x) dx,
 
where \int is the integration symbol. To actually calculate this with a real motorcycle would require numerical integration from a good estimate of that distribution of weight, which would be quite a project to obtain, probably using a dedicated jig and four scales. A well-chosen smooth function might do pretty well too, but the choice of a motorcycle density function isn't obvious, and would fit different motorcycles better or worse.

This whole exercise in the end shows that there is no universal solution for all motorcycles. It also suggests that my first model is a very poor guide. The fact that a motorcycle (not just the test bicycle) would balance pretty far away from 90 degrees means that the change in measured weight as the front is raised from the ground is on a steeper part of the cosine curve than model 1 had it. So the thickness of the scale would matter more than I first suggested.

In the end the best guide is probably the Aussie's empirical project with a real motorcycle that I recommended before

https://www.youtube.com/watch?v=_z7SUUYGRqU

At the end of his 10 minute demonstration the presenter is surprised to find the weight measurement was off by about 2 kg (not measured very precisely) when the front wheel was up on the scale. "Cool bananas!" he remarked. Not exactly as memorable as "Eureka," but, again, a very enjoyable experiment to watch.

Moto

[Huzo]

Pull it Dusty...

[swooshdave]

Because most bathroom scales don't go as high as 500 lbs.

I bet the ones at Walmart do.

[tris]

Bloody hell Moto - that's some hypothesis - my head now hurts!!!!

I think Muzz got it right on page one of this saga ......  pick the thing up and subtract your weight 

[fotoguzzi]

use a big fish scale?

[Tom]

That may be to simple. 

[Moto]

I looked at my earlier analysis and decided that the approach of calculating a torque for half the mass of the bike around a fixed pivot was, well, bogus.

So I did a satisfactory analysis by modeling the motorcycle as the abstraction seen at the top of this page:



All the mass (the total weight of the bike) is concentrated at point M, which is attached by a vertical bar to another bar that ends at the front and rear axles. Both bars are massless, and the axles are points. The angle beta is the angle between the center of mass M and the rear axle, which I measured by finding the point where the bike balances on its rear wheel. This, as mentioned before, is not achieved when the bike has been rotated up by 90 degrees, but instead when it has been rotated 19.1 degrees less, which is the numerical value of beta for the bicycle being used to model a motorcycle.

In part B the abstract motorcycle is elevated at an angle alpha above the horizontal. The applied force (weight) at the front axle F is equal to the total mass (M) times the ratio of the moment arm a = MN divided by the wheelbase, w. (This is a point mass model, instead of the distributed mass moment (integral) model proposed before.)

The page shows calculations of angle and line segment lengths using trigonometry and the given values, namely the wheelbase, w, the elevation, e, and the angle beta to the center of mass. The result is a single equation that gives the predicted weight on the front wheel as a function of its elevation:

             m3weight = M / (2 cos(beta)) * sin[pi/2 - beta - arcsin(elevation / wheelbase)]

The following graph shows how well the model does in reproducing the previously obtained measurements:



There are some interesting features to the graph. The heaviest dashed line represents the model 3 prediction, which is close to the observed values on the whole. It correctly predicts the observed weight when the elevation is zero, and also the correct front wheel weight of zero when the bike is at the balancing point [which a calculation from the measured angle (19.1 degrees) shows occurred at 703 mm of elevation of the front wheel]. The model doesn't do as well in predicting the weight when the front wheel is 248 mm below the horizontal (hanging off the end of the workbench, resting on the scale), for no reason that I can think of.

The most interesting part is model 3's predicted value when the bike is 90 degrees from horizontal, which is past the balancing point. The predicted weight on the front wheel is negative, about -1.6 kg. Of course I cannot directly measure this negative weight with my scale.

This implication of the model can be understood by bringing to mind that the total measured weight on both wheels together must be equal to the motorcycle's actual weight (as measured previously in a basket, at 11382 g). If the front wheel has a negative weight, then the rear wheel must have a measured weight that is greater than the motorcycle's actual basket weight!

This is easily confirmed by locking the rear wheel, placing the bike on the scale and rotating it to 90 degrees, which takes some forward-directed pressure from one's hand to keep the bike in position, of course. Doing this as carefully as I could, the reading on the scale was about 12.9 kg, or 1.6 kg greater than the bike's actual weight, even though I was very careful not to add any downward pressure.

Where does the extra apparent weight on the scale come from? It must come from my forward-directed pressure keeping the bike from falling backward. Newton's third law says that every force (my hand pressure) is countered by an opposite and equal force. The only available point on the bike through which that opposing force can be applied is at the tire's contact patch, sitting on the scale, and that's where it is coming from. There is a vector analysis that someone who knows what they're doing could perform to make it all balance out, but I could feel I was exerting about the right amount (1.6 kg, perhaps) of pressure with my hand, and I could see the scale measure the extra 1.6 kg of counteracting force in the difference between the apparent and true weights. My son was somewhat impressed, for once.   

In any event, this new model is a good one. It reproduces the data, and makes out-of-sample predictions that are confirmed. I plotted the extra 1.6 kg of apparent weight as its corresponding negative value in the plot, at the red triangle. This is an actual, indirect measurement of the counter-intuitive "negative weight" prediction of the model.

The upshot is still the same as before. Using a scale under one wheel will produce more of an error in calculating the total weight (as the sum of both wheels separately measured) than I figured from model 1, a simple, erroneous application of my  so-called Sine/Cosine Theory. Again, this is because the action is occurring in a steeper region of a trigonometric function (now represented as a sine). The error is probably in the range of a pound or two, I would guess, though I don't have any actual motorcycle measurements to base the guess on. The Australian still provides the best empirical evidence.

Moto

[Tom H]

With tongue  in cheek,

Unless your after weighing the bike for a racing series to make sure you meet the weight. Just put the bloody bikes front wheel on a scale while holding it upright, write it down and then do the same for the rear. Add the two together and it will tell you about what the heck it should weigh.

Wow, I never knew the math involved in this

Tom

[Tom]

Easier to take it to FEDEX at an airport and put it on the scale.  This could be too easy.    For the Daytona RS, I took the owner's manual and the DMV took the weight from the manual.