I have a delimea my motor wont turn start it turns overs, but wont start. pulled the plug and it has a weak spark that seems to weaken after each crank. I'm thinking its the battery but I cant test myself, multi meter dead. The starter also seems to slowing after a few cranks I charged the battery over night but its still doing the same thing. does this sound like the battery to anyone else or something else.  Thanks

With a bad battery, you'll be able to turn over but there won't be any juice left for sparking.

You can try push starting.

That may have been the easiest thread yet  

well when you turn it over the batter is going to drop more.

11 volts says at least 1 cell is dead, maybe 2.

FYI: from Wikipedia

The following are general voltage ranges for six-cell lead-acid batteries:

Open-circuit (quiescent) at full charge: 12.6 - 12.8 V
Open-circuit at full discharge: 11.8 - 12.0 V
Loaded at full discharge: 10.5 V.
Continuous-preservation (float) charging: 13.8V for gelled electrolyte; 13.5V for AGM (Absorbed Glass Mat) and 13.4V for flooded
All voltages are at 20C, and must be adjusted for temperature changes.
Float voltage recommendations vary, according to the manufacturers' recommendation.
Precise (+/- 0.05V) float voltage is critical to longevity; too low (sulfation) is almost as bad as too high (corrosion & electrolyte loss)
Typical (daily) charging: 14.2 - 14.5 V (depending on manufacturer's recommendation)
Equalization charging (for flooded lead acids): 15 - 16 V
Gassing threshold: 14.4 V
After full charge the terminal voltage will drop quickly to 13.2 V and then slowly to 12.6 V.

To add to the thread...

You could have a "good" battery that is just not fully charged at 10v and still start the bike.

You could have a bad battery at 13.6v and it won't....

Remember batteries are chemical reactions. On common car & bike batteries,  Lead plates sitting in acid.  the chemical reaction creates some volts.   Like Max says, colder less voltage, because cold slows down the reaction.

Common ways batteries wear out:

-  Shorting of the plates which is caused by the Byproduct of the chemical reaction.  Crud builds up on the bottom from the acid eating away the lead.

- Corrosion,  the lead can get covered by corrosion/oxidation if they get discharged too many times.  This causes the battery to deliver less current, but it can have same voltage.  Remember Power (wattage) runs things and power = voltage x current.  If you have voltage but little current, it won't go.  

Savage_Greg wrote on 12/31/69 at 16:00:11:

 Yea,  I went round and round wondering if corrosion was the right word to use.  Added Oxidation only because I couldn't remember what was in the acid.  But we add water to the battery and it's got O2 in it.  But.....just looked in a dictionary and corrorsion is  " wear away by chemical action" ...  hmmmmmmmm......

From VT engineering dept.  "The electrical force produced by an electrochemical cell is measured by the cell voltage. Cell voltage depends on the redox reactions occuring in the cell and the concentration of the reactants, but not on the number of electrons passing through the cell: it is an intensive property. The standard voltage E0 of a cell is measured when all reactants are at 25oC and at 1M concentration or 1 atm pressure.

For a copper-zinc cell like

Zn|Zn+2||Cu+2|Cu
the standard cell voltage E0 = 1.101 V.
Since we can split a redox reaction into two parts, we can also define standard voltages for both the oxidation and reduction parts of the reaction, Eox0 and Ered0. We arbitrarily pick the hydrogen reduction half reaction

2H+(aq) + 2e- -> H2(g)
to have Ered0 = 0, and measure all other half reaction voltages in relationship to it.
The oxidation half reactions are just the reaction run in reverse, and the half cell oxidation voltage is just the negative of the reduction voltage. For example:

Reduction: Zn+2(aq) + 2e- -> Zn(s)    Ered0 = -0.762 V
Oxidation: Zn(s) -> Zn+2(aq) + 2e-    Eox0 = 0.762 V
The standard voltage of a cell, E0, is just the sum of the standard voltages of the oxidation and reduction half reactions. This is independent of the number of electrons in the half reaction: if you need to multiply a half reaction by 3 to balance the redox equation, you do not multiply the cell voltage by 3. For example, the zinc-copper cell above is made up of the reactions

Oxidation: Zn(s) -> Zn+2(aq) + 2e-    Eox0 = 0.762 V
Reduction: Cu+2(aq) + 2e- -> Cu(s)    Ered0 = 0.339 V
E0 for the cell is Ered0 + Eox0 = 0.339 + 0.762 = 1.101 V
Example: What is the standard voltage for the cell

Mn|Mn+2||Fe+3|Fe+2
Solution: First, seperate the cell into two half reactions

Oxidation: Mn(s) -> Mn+2(aq) + 2e-
Reduction: Fe+3(aq) + e- -> Fe+2(aq)
The iron reduction has a voltage of 0.769 V. The oxidation of manganese is not listed, but the reduction (reverse reaction) is listed with a reduction voltage of -1.182 V. The oxidation thus has a voltage of +1.182 V
To compute the cell voltage, simply add the two voltages. It doesn't matter that the manganese reaction has two electrons and the iron only one

E0 = Ered0 + Eox0 = 0.769 + 1.182 = 1.951 V"

So, my 0.9-1.2 volt number was the oxidation portion of the reaction, didn't account for the reduction portion...