Renee' Turner/jennifer Bridges

10-05-1999, 09:10 AM

Thanks to all of the responses to my question.

Most of the responses were very similar so I've included this one as a

representative example:

The center of mass of the bicycle/rider system can be relatively easily

determined using a two step approach. First (the easy part), find the

bicycle

cg using a double suspension experimental method (suspension as in hanging,

not

as in shock absorbers). Hang the bicycle by one wheel from a ceiling hook,

beam

etc. You will probably have to tie the front wheel to the frame to keep it

in

line with the bicycle frame. Hang a plumb bob from the upper wheel

attachment

point so it hangs vertically along side the bicycle frame. The bicycle cg

is

along this line. Mark the intersection of this line on two points of the

frame

(say, the chainstay and the headtube) to establish this line for future use.

Now hang the bicycle again, using another attachment point (not the other

wheel). The seat post might be a good option. Repeat the plumb bob

procedure

with the bike in its new orientation (again, the plumb bob should be a

continuation of the suspension point towards the ground). Again, the bike

cg

lies along the plumb bob line. The intersection of the two lines

established

during the separate suspensions approximates the bicycle cg. The more

perpendicular these lines are to each other, the more accurate your

approximation of the bike cg.

Note that the horizontal cg of the bike can also be determined using a

reaction

board (knowing the bike weight, the bike position on the board, and the load

necessary to support one end of the board. Or the horizontal cg can be

determined by simply finding the point on the bike where a string can be

attached to suspend it in a perfectly level orientation. This approach also

works for the rider plus bike system, in the horizontal direction only of

course. But I find the double suspension system to be quick and easy, and

requires only some string, a small weight (for making a plumb bob), and a

ceiling hook.

Step two (the more complicated) is to find the cg of the rider - in the

riding

position. I refer you to a text by David Winter entitled Biomechanics of

Human

Movement for this procedure. Simply stated, you need to find the joint

locations, in two-dimensions (in the sagittal plane), describing the segment

orientations, locations, and lengths. Then, regression equations can be

used to

identify the center of mass locations for each segment. This can all be

done

manually using a good side view photograph of the rider and some grid

tracing

paper (digitize the joint centers, measure the separation distances defining

segments, determine center of mass locations within segments from regression

equations, and digitize centers of mass onto tracing paper). I assume you

do

not have motion capture and analysis systems which, from digitized video

images,

locate total body center of mass automatically.

The whole body cg (relative to one of the axles - for simplicity) is then

determined using the relation:

Xo = (m1x1 + m2x2 + m3x3 ..... mixi)/Mtotal

Yo = (m1y1 + m2y2 + m3y3 .....miyi)/Mtotal

where

Xo = horizontal location of body cg - relative to axle

Yo = vertical location of body cg - relative to axle

mi, m2, etc = mass of each segment (thigh, shank, arm, trunk, foot, etc.).

Don't forget two legs and arms.

x1, x2, ... xi = horizontal positions of centers of mass, relative to axle,

of

each segment (thigh, shank, foot, etc.)

y1, y2, ... yi = vertical positions of centers of mass, relative to axle, of

each segment

Mt = total mass of rider

Finally, the rider plus bike center of mass is determined from:

Xt = (mbxb + msxs)/Mbs

Yt = (mbyb + msys)/Mbs

where

Xt = horizontal location of rider plus bike cg, relative to axle

Yt = vertical location of rider plus bike cg, relative to axle

mb and ms = mass of bike and subject, respectively

xb, xs = horizontal position of bike cg and subject cg, relative to axle

yb, ys = vertical position of bike cg and subject cg, relative to axle

Mbs = the mass of the bicycle and subject combined

I hope this helps. The equations are easily put into a computer program

form.

In addition, the following discussion list "hardcore-bicycle science" was

also suggested:

http://www.sheldonbrown.com/hbs.html

Thanks again,

Jenni Bridges,

Michigan

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Most of the responses were very similar so I've included this one as a

representative example:

The center of mass of the bicycle/rider system can be relatively easily

determined using a two step approach. First (the easy part), find the

bicycle

cg using a double suspension experimental method (suspension as in hanging,

not

as in shock absorbers). Hang the bicycle by one wheel from a ceiling hook,

beam

etc. You will probably have to tie the front wheel to the frame to keep it

in

line with the bicycle frame. Hang a plumb bob from the upper wheel

attachment

point so it hangs vertically along side the bicycle frame. The bicycle cg

is

along this line. Mark the intersection of this line on two points of the

frame

(say, the chainstay and the headtube) to establish this line for future use.

Now hang the bicycle again, using another attachment point (not the other

wheel). The seat post might be a good option. Repeat the plumb bob

procedure

with the bike in its new orientation (again, the plumb bob should be a

continuation of the suspension point towards the ground). Again, the bike

cg

lies along the plumb bob line. The intersection of the two lines

established

during the separate suspensions approximates the bicycle cg. The more

perpendicular these lines are to each other, the more accurate your

approximation of the bike cg.

Note that the horizontal cg of the bike can also be determined using a

reaction

board (knowing the bike weight, the bike position on the board, and the load

necessary to support one end of the board. Or the horizontal cg can be

determined by simply finding the point on the bike where a string can be

attached to suspend it in a perfectly level orientation. This approach also

works for the rider plus bike system, in the horizontal direction only of

course. But I find the double suspension system to be quick and easy, and

requires only some string, a small weight (for making a plumb bob), and a

ceiling hook.

Step two (the more complicated) is to find the cg of the rider - in the

riding

position. I refer you to a text by David Winter entitled Biomechanics of

Human

Movement for this procedure. Simply stated, you need to find the joint

locations, in two-dimensions (in the sagittal plane), describing the segment

orientations, locations, and lengths. Then, regression equations can be

used to

identify the center of mass locations for each segment. This can all be

done

manually using a good side view photograph of the rider and some grid

tracing

paper (digitize the joint centers, measure the separation distances defining

segments, determine center of mass locations within segments from regression

equations, and digitize centers of mass onto tracing paper). I assume you

do

not have motion capture and analysis systems which, from digitized video

images,

locate total body center of mass automatically.

The whole body cg (relative to one of the axles - for simplicity) is then

determined using the relation:

Xo = (m1x1 + m2x2 + m3x3 ..... mixi)/Mtotal

Yo = (m1y1 + m2y2 + m3y3 .....miyi)/Mtotal

where

Xo = horizontal location of body cg - relative to axle

Yo = vertical location of body cg - relative to axle

mi, m2, etc = mass of each segment (thigh, shank, arm, trunk, foot, etc.).

Don't forget two legs and arms.

x1, x2, ... xi = horizontal positions of centers of mass, relative to axle,

of

each segment (thigh, shank, foot, etc.)

y1, y2, ... yi = vertical positions of centers of mass, relative to axle, of

each segment

Mt = total mass of rider

Finally, the rider plus bike center of mass is determined from:

Xt = (mbxb + msxs)/Mbs

Yt = (mbyb + msys)/Mbs

where

Xt = horizontal location of rider plus bike cg, relative to axle

Yt = vertical location of rider plus bike cg, relative to axle

mb and ms = mass of bike and subject, respectively

xb, xs = horizontal position of bike cg and subject cg, relative to axle

yb, ys = vertical position of bike cg and subject cg, relative to axle

Mbs = the mass of the bicycle and subject combined

I hope this helps. The equations are easily put into a computer program

form.

In addition, the following discussion list "hardcore-bicycle science" was

also suggested:

http://www.sheldonbrown.com/hbs.html

Thanks again,

Jenni Bridges,

Michigan

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To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

---------------------------------------------------------------